698 + Interview Questions in GEOMETRICAL-OPTICS IN GENERAL AWARENESS Page 11 InterviewSolution

501.	A lens if placed between a source of light and a wall. It forms images of area` A_(1)` and `A_(2)` on the wall for its two different positions. The area of the source or light isA. `(A_(1)+A_(2))/(2)`B. `[(1)/A_(1)+(1)/A_(2)]^(-1)`C. `sqrt(A_(1)A_(2))`D. `[(sqrt(A_(1))+sqrt(A_(2)))/(2)]^(2)`
Answer» Correct Answer - C `m_(1)=(A_(1))/(O)` and `m_(2)=(A_(2))/(O)` `impliesm_(1)m_(2)=(A_(1)A_(2))/(O_(2))` Also it can be proved that `m_(1)m_(2)=1` So `O=sqrt(A_(1)A_(2))`

Discussion

502.	The given lens is broken into four parts rearranged as shown. If the initial focal length is f, then after rearrangement the equivalent focal length is A. fB. `f//2`C. `f//4`D. `4f`
Answer» Correct Answer - b. Cutting a lens in transverse direction double their focal length, i.e., 2f. Using the formula of equivalent focal length, `(1)/(f)=(1)/(f_(1))+(1)/(f_(2))+(1)/(f_(3))+(1)/(f_(4))` We get equivalent focal length as `(f)/(2)` .

Discussion

503.	An object AB of height 1cm lying on the axis of convex mirror of focal length20cm, at a distance of 30cm from its pole as shown in figure. Now the object starts moving with a constant speed `1 cm//sec` towards the pole. Calculate Q. Magnitude of velocity of image at the given instantA. `0.08cms^(-1)`B. `0.64cms^(-1)`C. `0.16cms^(-1)`D. `0.32cms^(-1)`
Answer» Correct Answer - c. (i) Given : `u=-30cm, f=20cm` `(1)/(v)+(1)/(u)=(1)/(f)rArrv=12cm` Also, `(dv)/(dt)=-((v^(2))/(u^(2)))(du)/(dt)=-0.16cms^(-1)` (ii) `H=-(v)/(u)h`, H is the height of image and h is height of object. `(dH)/(dt)=(u(dv)/(dt)-v(du)/(dt))/(u^(2))h=0.008cms^(-1)`

Discussion

504.	An object AB of height 1cm lying on the axis of convex mirror of focal length20cm, at a distance of 30cm from its pole as shown in figure. Now the object starts moving with a constant speed `1 cm//sec` towards the pole. Calculate Q. Rate of change of height of image.A. `0.16cms^(-1)`B. `0.004cms^(-1)`C. `0.016cms^(-1)`D. `0.0008cms^(-1)`
Answer» Correct Answer - d. (i) Given : `u=-30cm, f=20cm` `(1)/(v)+(1)/(u)=(1)/(f)rArrv=12cm` Also, `(dv)/(dt)=-((v^(2))/(u^(2)))(du)/(dt)=-0.16cms^(-1)` (ii) `H=-(v)/(u)h`, H is the height of image and h is height of object. `(dH)/(dt)=(u(dv)/(dt)-v(du)/(dt))/(u^(2))h=0.008cms^(-1)`

Discussion

505.	The magnification of an object placed in front of a convex lens is `+2` . The focal length of the lens is `2.0` metres. Find the distance by which object has to be moved to obtain a magnification of `-2` (in metres).
Answer» Correct Answer - 2 For magnification `+2, u=-x, v=-2x ` and `f=2.0m` From `(1)/(v)-(1)/(u)=(1)/(f)`, we have `(1)/(-2x)+(1)/(x)=(1)/(z)` `x=1.0m` For magnification of `-2, u=-y, V=+2y, f=2.0m` we have `(1)/(2y)+(1)/(y)=(1)/(2)rArr y=3.0m` So distance to be moved `=y-x=3-1=2m`

Discussion

506.	The magnification of an object placed it front of a convex lens of focal length `20cm` is `+2`. To obtain a magnification of `-2` , the object will has to be moved a distance equal toA. `10 cm`B. `20 cm`C. `30 cm`D. `40cm`
Answer» Correct Answer - B When magnification is `+2`. `u=-x` `v=-2x` `f=+20` Using `(1)/(-2x)+(1)/(x)=(1)/(20)` or `x=10cm` To have a magnificatio of `-2` `u=-y` `v=+2y` and `f=+20` `:. (1)/(2y)+(1)/(y)=(1)/(20)` or `y=30cm` `:. y-x=20cm`

Discussion

507.	The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front of it to 50cm, the magnification of its image changes from `m_(25)` to `m_(50)`. The ration `m_(25)//m_(50)` is
Answer» Correct Answer - 6 m=(f)/(f+u)`

Discussion

508.	A combination of two thin lenses with focal lengths `f_(1)` and `f_(2)` respectively forms and image of distant object at distance `60cm` when lenses are in contact. The position of this image shifts by `30 cm` towards the combination when two lenses are separated by `10 cm`. The corresponding values of `f_(1)` and `f_(2)` areA. `30cm,-60cm`B. `20cm, -30 cm`C. `15cm, -20cm`D. `12 cm,-15cm`
Answer» Correct Answer - B `(1)/(60)=(1)/(f_(1))+(1)/(f_(2)),` `(1)/(30)=(1)/(f_(1))+(1)/(f_(2))-(10)/(f_(1)f_(2))` or `(1)/(30)=(1)/(60)-(10)/(f_(1)f_(2))` or `(10)/(f_(1)f_(2))=(1)/(60)-(1)/(30)` or `(10)/(f_(1)f_(2))=(1-2)/(60)=-(1)/(60)`or `f_(1)f_(2)=-600` or `f_(2)=(600)/(f_(1))` From (1) , `(1)/(60)=(1)/(f_(1))+(1)/((-600)/(f_(1)))` or`(1)/(60)=(1)/(f_(1))-(f_(1))/(60)` or `(1)/(60)=(60-f_(1)^(2))/(600f_(1))` or `600f_(1)=36000-60f_(2)^(1)` or `60g_(1)^(2)+600f_(1)-36000=0` or `f_(1)^(2)+10f_(1)-600=0` or`f_(1)^(2)+30f_(1)-20f_(1)-600=0` or `f_(1)(f_(1)+30)-20(f_(1)+30)=0` `f_(1)=20cm `or `f_(1)=-30cm` Now, `f_(2)=(600)/(20)cm` or `(-600)/(-30)` or`f_(2)=-30cm or 20cm`

Discussion

509.	A combination of two thin lenses with focal lengths `f_(1)` and `f_(2)` respectively forms and image of distant object at distance `60cm` when lenses are in contact. The position of this image shifts by `30 cm` towards the combination when two lenses are separated by `10 cm`. The corresponding values of `f_(1)` and `f_(2)` areA. `30cm, -60 cm`B. `20cm, -30 cm`C. `15 cm, -20cm`D. `20 cm, -15 cm`
Answer» Correct Answer - B `(1)/(60)=(1)/(f_(1))+(1)/(f_(2))` (i) and `(1)/(30)=(1)/(f_(1))+(1)/(f_(2))-(10)/(f_(1)f_(2))` (ii) On solving (i) and (ii) `f_(1)f_(2)=-600` and `f_(1)+f_(2)=-10` Hence`f_(1)=20cm` and `f_(2)= -30cm`

Discussion

510.	Two convex lenses of focal lengths `f_(1) and f_(2)` are placed coaxially, a distance `d` apart. If the axis of one of the lenses is lifted parallel to itself by `Delta`, find the distance by which the focal point is shifted and the distance of the focal point from the first lens.
Answer» Correct Answer - `(Delta(f_(1)-d))/(f_(1)+f_(2)-d); (f_(1)f_(2)+d(f_(1)-d))/(f_(1)+f_(2)-d)`

Discussion

511.	In a laboratory four convex lenses `L_(1), L_(2), L_(3)` and `L_(4)` of focal lengths `2,4,6` and `8 cm` respectively are available. Two of these lenses form a telescope of length `10 cm` and magnifying power 4 . The objective and eye lenses areA. `L_(2),L_(3)`B. `L_(1),L_(4)`C. `L_(3),L_(2)`D. `L_(4),L_(1)`
Answer» Correct Answer - D To make telescope of higher magnifying power, `f_(o)` should be large `f_(e)` should be least.

Discussion

512.	An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation ` A. `PQ` is horizontalB. `QR` is horizontalC. `RS` is horizontalD. Either `PQ` or `RS` is horizontal
Answer» Correct Answer - B In minimum deviation position refracted ray inside the prism is parallel to the base of the prism.

Discussion

513.	An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minumum deviation, A. PQ is horizontalB. QR is horizontalC. Rs is horizontalD. any one will be horizontal
Answer» Correct Answer - b. For minimum deviation, incident angle is equal to emerging angle. Therefore, QR is horizontal.

Discussion

514.	A point object is placed at the centre of a glass sphere of radius 6cm and refractive index 1.5. The distance of virtual image from the surface isA. 2 cmB. 4 cmC. 6 cmD. 12 cm
Answer» Correct Answer - C When the object is placed at the centre fo the glass sphere, the rays from the sphere and emerge undeviated .

Discussion

515.	A point object is placed at the centre of a glass sphere of radius 6cm and refractive index 1.5. The distance of virtual image from the surface isA. `2cm`B. ` 4 cm`C. `6 cm`D. `12 cm`
Answer» Correct Answer - C Using refractio formula `(._(1)mu_(2)-1)/(R)=(._(1)mu_(2))/(v)-(1)/(u)` in given case, medium `(1)` in glass and `(2)` is air So `(._(g)mu_(a)-1)/(R)=(._(g)mu_(a))/(v)-(1)/(u)implies((1)/(1.5)-1)/(-6)=(1)/(1.5v)-(1)/(-6)` `=(1-1.5)/(-6)=(1)/(v)+(1.5)/(6)implies(0.5)/(6)=(1)/(v)+(1)/(4)` `implies(1)/(v)=(1)/(12)-(1)/(4)=-(2)/(12)=-(1)/(6)implies6 cm.`

Discussion

516.	Two absolute scales `X` and `Y` assigned numerical values `200` and `450` to the triple of water. What is the relation between `T_(X) ` and `T_(Y)` ?A. `9T_(X)= 4T_(Y)`B. `4T_(X)= 9T_(Y)`C. `T_(X) = 3T_(Y)`D. None of these
Answer» Correct Answer - A If we take two fixed points as tripe point of water and `0K`. Then `(T_(X)-0)/(200) = (T_(Y)-0)/(450) implies 450 T_(X) = 200. T_(Y) implies 9T_(X) = 4T_(Y)`

Discussion

517.	A meter washer has a hole of diameter `d_(1)` and external diameter `d_(2)`, where `d_(2)=3d_(1)`. On heating , `d_(2)` increases by `0.3%`. Then `d_(1)` will :-A. decreases by ` 0.1 %`B. decrease by `0.3%`C. increases by `0.1%`D. increase by `0.3%`.
Answer» Correct Answer - D Expansion of a metal is same as photographic enlargement `implies d_(1)` will increase by `0.3%` `implies d_(1) , 0.3%`

Discussion

518.	A faulty thermometer reads freezing point and boiling point of water as `-5 ^(@)C` abd `95^(@)C` respectively . What is the correct value of temperature as it reads `60^(@)C` on faulty thermometer?A. `60^(@)C`B. `65^(@)C`C. `64^(@)C`D. `62^(@)C`
Answer» Correct Answer - B `(X-LFP)/(UPF-LFP)` = constant ( for all temperature scales ) where `LFP rarr` lower fixed point `UFP rarr` Upper fixed point `(X-(-5))/(95-(-5)) = (C-0)/(100-0) implies (60+5)/(95+5) = (C)/(100) implies C= 65^(@)C`

Discussion

519.	A steel scale is to be prepared such that the millimeter intervals are to be accurate within `6 xx10^(-5)mm`. The maximum temperature variation form the temperature of calibration during the reading of the millimeter marks is `(alpha = 12 xx 10^(-6)//"^(@)C)`A. `4.0^(@)C`B. `4.5^(@)C`C. `5.0^(@)C`D. `5.5^(@)C`.
Answer» Correct Answer - C `DeltaL= 6xx10^(-5) = Lalphatheta implies theta = (6xx10^(-5))/(1 xx 12 xx 10^(-6) =5^(@)C`

Discussion

520.	A water cooler of storages capacity 120 liters can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3kW of heat (thermal load). The temperature of water fed into the device cannot exceed `30^@C` and the entire stored 120 liters of water is initially cooled to `10^@C.` The entire system is thermally insulated. The minimum value of P ( in watts) for which the device can be operated for 3hours is (Specific heat of water is `4.2kJkg^-1K^-1` and the density of water is `1000kgm^-3`)A. `1600`B. `2067`C. `2533`D. `3933`
Answer» Correct Answer - B `3000-P = (120xx1)(4.2xx10^(3))(dT)/(dt)` `(dT)/(dt) = (20)/(60 xx 60xx 3)` `P = 2067W`

Discussion

521.	A convex lens of focal length 20cm and a concave lens of focal length f are mounted coaxially 5cm apart. Parallel beam of light incident on the convex lens emerges from the concave lens as a parallel beam. Then, f in cm isA. 35B. 25C. 20D. 15
Answer» Correct Answer - d. Clearly, power of system is zero. `:.0=(1)/(20)+(1)/(f)-(5)/(20f)` or `-(1)/(20)=(15)/(20f) or f=-15cm`

Discussion

522.	A convex lens A of focal length 20cm and a concave lens G of focal length 5cm are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance d in cm will beA. 25B. 15C. 30D. 50
Answer» Correct Answer - b. `P=(1)/(f_(1))+(1)/(f_(2))-(d)/(f_(1)f_(2))` `0=(1)/(20)-(1)/(5)-(d)/(20(-5))` `(d)/(100)=(1)/(5)-(1)/(20)=(4-1)/(20)=(3)/(20)` or `d=15cm`

Discussion

523.	A convex lens forms an image of an object placed 20cm away from it at a distance of 20cm on the other side of the lens. If the object is moves 5cm toward the lens, the image will beA. `5 cm` towards the lensB. `5 cm` away from the lensC. `10 cm` towards the lensD. `10 cm` away from the lens
Answer» Correct Answer - D Clearly, `2f=20cm` or `=10cm` Now, `u=-15cm, v=?` `f=10cm` `(1)/(v)-(1)/(-15)=(1)/(10)` or `(1)/(v)+(1)/(15)=(1)/(10)` or `(1)/(v)=(1)/(10)-(1)/(15)`. or `(1)/(v)=(3-2)/(30)=(1)/(30)` or `v=30cm` The change in image distance is `(30-20)` i.e., `10 cm`.

Discussion

524.	Figure, shows variation of magnification m (produced by a thin convex lens) and distance v of image from pole of the lens. When of the following statements are correct? A. Focal length of the lens is equal to intercept on v-axisB. Focal length of the lens is equal to inverse of slope of the lineC. Magnitude of intercept on m-axis is equal to unityD. None of the above
Answer» Correct Answer - a.,b.,c. `(1)/(v)-(1)/(u)=(1)/(f)` `1-(v)/(u)=(v)/(f)` (i) Since m is positive in the given figure, therefore m represents magnitude of the magnification. In fact, if a real image is formed by a convex lens, then the image and object will be on opposite sides of the lens. It means, if v is positive, then u will be negative . Therefore, `m=\|(v)/(u)\|=-(v)/(u)` (ii) Hence, Eq. (i) becomes `m=v/(f)-1` It means, the graph between m and v will be a straight line having intercept `-1` on m-axis and slope of the ling `tantheta` is equal to `(1//f)`. Hence, option (b) and (c) are correct. Putting `m=0` in Eq. (ii), `v=f` . Hence (a) is also correct. Obviously, (d) is wrong.

Discussion

525.	The graph shows how the magnification `m` produced by a convex thin lens varies with image distance `v`. What was the focal length of the used ? A. `(b)/(c)`B. `(b)/(ca)`C. `(bc)/(a)`D. `(c)/(b)`
Answer» Correct Answer - D For a lens `m=(f-v)/(f)=-(1)/(f)v+1` Comparing it with `y=mx+c` Slope `=m=-(1)/(f)` From graph, slpe of the line `=(b)/(c)` Hence `-(1)/(f)=(b)/(c)implies\|f\|=(c)/(b)`

Discussion

526.	A convergent beam of light converges to a point `20 cm` behind the convex mirrorr on the principal axis.An inverted image of the same size is formed coincident with the virtual object. Then, the focal length of the convex mirrorr isA. `20 cm`B. `10 cm`C. `40 cm`D. `30 cm`
Answer» Correct Answer - B `(1)/(20)+(1)/(20)=(1)/(f)` or`(1)/(f)=(1)/(10)` or`f=10cm`.

Discussion

527.	The distance between an electric lamp and a screen is `d=1m`. A convergent lens of focal length `f=21`cm is placed between the lamp and the lens such that a sharp image of the lamp filament is formed on the screen.A. The positions of the lens from the lamp for which sharp images are formed on the screen are 35cm and 65cm.B. The positions of the lens from the lamp for which sharp images are formed on the screen are 35cm and 70cmC. Magnitude of the difference in magnification is `40//21`D. The size of the lamp filament for which there are two sharp images of 4.5cm and 2cm, is 3cm
Answer» Correct Answer - b.,c. `(1)/(100-x)=(1)/(-x)=(1)/(21)` `rArr` x, The distance of object from the lens is 30cm, 70cm `m_(1)=(70)/(-30),m_(2)=(30)/(-70):.\|m_(1)-m_(2)\|=(40)/(21)`

Discussion

528.	A diatomic gas is enclosed in a vessel fitted with massless movable piston. Area of cross section of vessel is `1m^2`. Initial height of the piston is 1m (see the figure). The initial temperature of the gas is 300K. The temperature of the gas is increased to 400K, keeping pressure constant, calculate the new height of the piston. The piston is brought to its initial position with no heat exchange. Calculate the final temperature of the gas. You can leave answer in fraction.
Answer» Correct Answer - `400((4)/(3))^(0.4) K` At constant pressure `V alpha T implies (V_(2))/(V_(1)) = (T_(2))/(T_(1)) implies (Ah_(2))/(Ah_(1)) = (T_(1))/(T_(1))` `therefore h_(2) = h_(1) ((T_(2))/(T_(1))) = (1.0) ((400)/(300))m = (4)/(3)m` As there is no heat loss , process is adiabatic. For adiabatic process `T_(f)V_(f)^(gamma-1) = T_(t)V_(t)^(gamma-1)` `therefore T_(f) = T_(t)((V_(t))/(V_(f)))^(gamma-1) = (400)((h_(t))/(h_(f)))^(1.4-1) = 400((4)/(3))^(0.4)`

Discussion

529.	An image of a bright square is obtained on a screen with the aid of a convergent lens. The distance between the square and the lens is 40cm. The area of the image is nine times larger than that of the square. Select the correct statement(s):A. Image is formed at a distance of 120cm from the lensB. Image is formed at a distance of 360cm from the lensC. Focal length of the lens is 30cmD. Focal length of the lens is 36cm
Answer» Correct Answer - A::C `(v)/(u)=(3)/(1)rArrv=3u` `(1)/(v)-(1)/(u)=(1)/(f),(1)/(3u)+(1)/(u)=(1)/(f)` `f=30cm.`

Discussion

530.	A Galileo telescope has an objective of focal length `100cm` and magnifying power 50. The distance between the two lenses in normal adjustment will beA. `98 cm`B. `100cm`C. `150cm`D. `200 cm`
Answer» Correct Answer - A `m=(f_(o))/(f_(e))implies(100)/(f_(e))=50impliesf_(e)=2cm` Normal distance `f_(o)-f_(e)=100-2=98cm`

Discussion

531.	A point object is located at a distance of 100cm from a screen. A lens of focal length 23 cm mounted on a movable frictionless stad is kept between the source and the screen. The stand is attached to a spring of natureal length 50cm and spring constant `800N//m` as shown in Figure . Mass of the stand with lens is 2 kg. How much impulse P should be imparted to the stand so that a real image of the object is formed on the screen after a fixed time gap? Also find this time gap. (Neglect width width of the stand)
Answer» Correct Answer - `8km//s` Let the difference of th elens from the object be `l` when a real image is formed on the screen. Then, `(1)/(100-l)-(1)/(-l)=(1)/(23)` On solving, we get `l=(50+-sqrt(2))cm` Now, if the lens performs SHM and real image is formed after a fixed time gap then this time gap must be one fourth of the time period. Therefore, phase difference between the two positions of real image must be `(pi)/(2)` . As the two positions are symmetrically located about the origin, phase difference of any of these positions from origin must be `(pi)/(4)`. `rArr 10sqrt(2)cm=A "sin" (pi)/(5)rArr A=20cm` To achieve this velocity at the mean position, `v_(0)=A_(omega)=Asqrt((K)/(m))` `:.` Required impulse `rho=mv_(0)=Asqrt(Km)=8kms^(-1)`

Discussion

532.	A thin equiconvex glass lens `(mu_(g)=1.5)` is beign placed on the top of a vessel of height `h=20cm` as shown figure. A luminous point source is beign placed at the bottom of vessel on the principal axis of the lens. When the air is on both the side of the lens the image of luminous source is formed at a distance of 20cm from the lens out side the vessel. When the air inside the vessel is being replaced by a liquid of refractive index `mu_(l)` the image of the same source is being formed at a distance 30cm from the lens outside the vessel. Find the `mu_(l)`.
Answer» Correct Answer - 1.11 For the first lens, `(1)/(v)-(1)/(u)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))` `(1)/(20)-(1)/(-20)=(1.5-1)((1)/(R)-(1)/(-R))rArrR=10cm` For the second case when the vessel is being fillle dwith the liquid then `(mu_(air))/(v_(1))-(mu_(l))/(u)=(u_(g)-mu_(l))/(R)+(mu_(air)-mu_(g))/(-R)` `(1)/(30)-(mu_(l))/(-20)=(1.5-mu_(l))/(10)+(1-1.5)/(-10)` `(mu_(l))/(20)+(mu_(l))/(10)=(3)/(20)+(1)/(20)-(1)/(30)` `(3mu_(l))/(20)=(1)/(6)` `mu_(l)=(10)/(9)=1.11`

Discussion

533.	A thin equiconvex glass lens `(mu_(g)=1.5)` is beign placed on the top of a vessel of height `h=20cm` as shown figure. A luminous point source is beign placed at the bottom of vessel on the principal axis of the lens. When the air is on both the side of the lens the image of luminous source is formed at a distance of 20cm from the lens out side the vessel. When the air inside the vessel is being replaced by a liquid of refractive index `mu_(l)` the image of the same source is being formed at a distance 30cm from the lens outside the vessel. Find the `mu_(l)`.
Answer» Correct Answer - 1.11

Discussion

534.	A person can see objects clearly only when they lie between `50cm` and `400 cm` from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will beA. concave, `-0.2`diopterB. convex, `+0.15` diopterC. convex,`+2.25` diopterD. concave, `-0.25`diopter
Answer» Correct Answer - D As we want to correct myopia , so, far point must go to infinity. `u=-4m, u=-oo,P=?` `P=(1)/(f)=(1)/(v)-(1)/(u)=(1)/(-4)-(1)/(oo)=-0.25D` `(-)` implies concave mirrorr

Discussion

535.	A glass `(mu = 1.5)` sphere of radius R is viewed from outside along a diameter. Calculate the distance between two points (say P and Q) lying on the line AB whose images are seen at centre C and point A respectively.
Answer» Correct Answer - `(R)/(2)`

Discussion

536.	A large rectangular glass block of refractive index `mu` is lying on a horizontal surface as shown in Figure. Find the minimum value of `mu` so that the spot S on the surface cannot be seen through top plane ‘ABCD’ of the block.
Answer» Correct Answer - `mu_(min) = sqrt(2)`

Discussion

537.	The focal length of the lens used in question 118 isA. `(D^(2)+d^(2))/(2D)`B. `(D^(2)-d^(2))/(4D)`C. `(D^(2)-d^(2))/(2D)`D. `(D^(2)+d^(2))/(d)`
Answer» Correct Answer - b. As `d=sqrt(D(D-4f))` `d^(2)=D^(2)-4Df` `f=(D^(2)-d^(2))/(4D)`

Discussion

538.	Object O is placed in front of a plane mirror M. A glass slab S having thickness t = 3 cm and refractive index `mu = 1.8` is placed between the object and the mirror. The refracting faces of the slab are parallel to the mirror surface. The object is made to move with a velocity of `u = 2 m//s` perpendicular to the mirror surface. Find the speed of the image formed after (a) refraction from the slab followed by reflection from the mirror. (b) refraction from the slab followed by reflection from the mirror followed by the refraction from the slab.
Answer» Correct Answer - (a) `2m//s`

Discussion

539.	A point object (A) is kept at a distance (L) from a convex mirror on its principal axis. A glass slab is inserted between the object and the mirror with its refracting surfaces perpendicular to the principal axis of the mirror. The thickness of the slab is 6 cm and its refractive index is `(3)/(2)`. The image formed after refraction through the slab, reflection from the mirror followed by refraction through the slab is a virtual image at a distance of 10 cm from the pole of the mirror (on its principal axis). Consider paraxial rays only and calculate the distance (L) of the object from the mirror. Focal length of the mirror is `f = 20 cm`.
Answer» Correct Answer - `L = 32 cm`

Discussion

540.	This question concerna a symmetrical lens shown, along with its two focal points. It is made of plastice with `n=1.2` and has focal length f. Four different regions are shown: Here, A. `-oox lt -f` B. `-f lt x lt 0` C. `0 lt x lt f` D. `f lt x lt oo` Q. A second lens is now placed adjacent to the first in contact with it. The new lens is concave, with a focal length `f=-3f`. What is the focal length of the two-lens system? Treat the lenses as thin.A. AB. BC. CD. D
Answer» Correct Answer - d From the formula of equivalent focal length of thin lenses in contact, we get `(1)/(F)=(1)/(f_(1))+(1)/(f_(2))=(1)/(f)-(1)/(3f) or F=(3f)/(2)`

Discussion

541.	A point object O is placed at a distance of 62 cm in front of a concave mirror of focal length`f = 20 cm`. A glass slab of refractive index `mu = (3)/(2)` and thickness 6 cm is inserted between the object and the mirror. Let’s define final image as image formed after the light ray originating from O passes through the slab, gets reflected from the mirror and then again passes through the slab. At what distance d from the mirror, the face AB of the slab can be placed so that the final image is formed inside the slab itself?
Answer» Correct Answer - `26 cm lt d lt 32 cm`

Discussion

542.	Figure shows a torch producing a straight light beam falling on a plane mirror at an angle `60^(@)` The reflected beam makes a spot P on the screen along y-axis . If at t=0, mirror starts ratating about the hinge A with an angular velocity `(omega)=1^(@)` per second clockwise. Find the speed of the spot on screen after time t = 15 s.
Answer» Correct Answer - `(2pi)/(15)m//s`

Discussion

543.	Can we project the image formed by a plane mirror on to a screen? Give reasons.
Answer» Yes, for a virtual object. Because the plane mirror will form a the real image of a virtual object and real image can be takne on a screen.

Discussion

544.	A ray of light travelling in the direction `(1)/(2)``(hati,+sqrt3hatj)` is incident on a plane mirror. After reflection, it travels along the direction (1)/(2)` `(hati-sqrt3hatj)` . The angle of incidence isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `75^(@)`
Answer» Correct Answer - A Here normal is along `hatj` Angle between incident ray and normal `"cos"theta = ((1)/(2) (hati + sqrt(3hatj)). Hatj)/((1)(1)) = (sqrt(3))/(2) implies theta = 30^(@)`

Discussion

545.	A ray of light in incident along a vector `hati + hatj - hatk` on a plane mirror lying in y-z plane. The unit vector along the reflected ray can beA. `(hati + hatj -hatk)/(sqrt(3))`B. `(hati - hatj +hatk)/(sqrt(3))`C. `(-hati + hatj -hatk)/(sqrt(3))`D. `(3hati + hatj -hatk)/(sqrt(3))`
Answer» Correct Answer - C::D According to law of reflection `hatr= hate -2(hate.hatn)hatn "Here" hate= (hati + hatj -hatk)/(sqrt(3)), hatn =pmhati` so `hate.hatn = (1)/(sqrt(3)) implies hatr = (hati + hatj - hatk)/(sqrt(3))pm (2hati)/(sqrt(3)) = (3hati + hatj - hatk)/(sqrt(3)) "or" (-hati + hatj -hatk)/(sqrt(3))`

Discussion

546.	A ray of light incident normally on one of the faces of a right-angle isosceles prism is found to be totally reflected as shown in the figure. What is the minimum value of the refractive index of the material of the prism? When the prism is immersed in water, trace the path of the emergent ray for the same incident ray, indicating the values of all the angles `(mu_omega=4//3)`.
Answer» Correct Answer - `1.414, 48.6^(@)`

Discussion

547.	A ray reflected successively from two plane mirrors inclined at a certain angle `(lt 90^(@))` undergoes a deviation of `300^(@)` . The number of images observable are:A. 10B. 11C. 12D. 14
Answer» Correct Answer - B If an angle between mirrors is `theta` then the deviation from two plane mirrors will be =` 2pi -2 theta` `300 = 2pi - 2 theta implies theta = 30^(@)`, Hence `(300)/(30)= 12` Number of images = m-1 = 12-1 = 11

Discussion

548.	An air bubble in a glass sphere `(mu = 1.5)` is situated at a distance `3 cm` from a convex surface of diameter `10 cm`. At what distance from the surface will the bubble appear ?
Answer» In case of refraction from curved surface `(mu_(2))/(v) -(mu_(1))/(u) = ((mu_(2)-mu_(1))/(R)` (a) " " `mu_(1) = 1.5 , mu_(2) = 1 , R = -5"cm" "and" u=-3 "cm" implies (1)/(v)- ((1.5))/((-3)) = (1-1.5)/((-5)) implies v=-2.5 "cm"` the bubble will appear at a distance `2.5"cm"` from the convex curved surface inside the glass. (b) `mu_(1) = 1.5 , mu_(2) =1 , R =5 "cm" "and" u = -3 "cm" implies (1)/(v)- ((1.5))/((-3)) = (1-1.5)/((5)) implies v= -1.66 "cm"` the bubble will appear at a distance `1.66` cm from the concave curved surface inside the glass . Note: If the surface is plane then `R rarr oo` case(a) or (b) would yield `(1)/(v) - ((1.5))/((-3)) = ((1 - 1.5))/(oo) implies v= -2"cm"`

Discussion

549.	An equiconvex lens has its two surfaces of radius of curvature `R = 10 cm`. Thickness of the lens at its centre is 2 cm. A light ray is incident making an angle of `2°` with the optic axis of the lens. Find the angle that the emergent ray will make with the optic axis. The refractive index of all media is as marked in the Figure.
Answer» Correct Answer - `((8)/(15))^(@)`

Discussion

550.	An equiconvex lens of refractive index 1.5 has its two surfaces having radius of curvature of 30 cm. A point object has been placed on the principal axis at a distance of 60 cm from the lens. Find the distance of image from the lens formed by the rays which suffer refraction at first surface, reflection at second surface, again a reflection on the first surface and finally a refraction from the second surface.
Answer» Correct Answer - `(60)/(13)cm`

Discussion

Explore topic-wise InterviewSolutions in .

A lens if placed between a source of light and a wall. It forms images of area` A_(1)` and `A_(2)` on the wall for its two different positions. The area of the source or light isA. `(A_(1)+A_(2))/(2)`B. `[(1)/A_(1)+(1)/A_(2)]^(-1)`C. `sqrt(A_(1)A_(2))`D. `[(sqrt(A_(1))+sqrt(A_(2)))/(2)]^(2)`

The given lens is broken into four parts rearranged as shown. If the initial focal length is f, then after rearrangement the equivalent focal length is A. fB. `f//2`C. `f//4`D. `4f`

The magnification of an object placed in front of a convex lens is `+2` . The focal length of the lens is `2.0` metres. Find the distance by which object has to be moved to obtain a magnification of `-2` (in metres).

The magnification of an object placed it front of a convex lens of focal length `20cm` is `+2`. To obtain a magnification of `-2` , the object will has to be moved a distance equal toA. `10 cm`B. `20 cm`C. `30 cm`D. `40cm`

The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front of it to 50cm, the magnification of its image changes from `m_(25)` to `m_(50)`. The ration `m_(25)//m_(50)` is

Two convex lenses of focal lengths `f_(1) and f_(2)` are placed coaxially, a distance `d` apart. If the axis of one of the lenses is lifted parallel to itself by `Delta`, find the distance by which the focal point is shifted and the distance of the focal point from the first lens.

An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation ` A. `PQ` is horizontalB. `QR` is horizontalC. `RS` is horizontalD. Either `PQ` or `RS` is horizontal

An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minumum deviation, A. PQ is horizontalB. QR is horizontalC. Rs is horizontalD. any one will be horizontal

A point object is placed at the centre of a glass sphere of radius 6cm and refractive index 1.5. The distance of virtual image from the surface isA. 2 cmB. 4 cmC. 6 cmD. 12 cm

A point object is placed at the centre of a glass sphere of radius 6cm and refractive index 1.5. The distance of virtual image from the surface isA. `2cm`B. ` 4 cm`C. `6 cm`D. `12 cm`

Two absolute scales `X` and `Y` assigned numerical values `200` and `450` to the triple of water. What is the relation between `T_(X) ` and `T_(Y)` ?A. `9T_(X)= 4T_(Y)`B. `4T_(X)= 9T_(Y)`C. `T_(X) = 3T_(Y)`D. None of these

A meter washer has a hole of diameter `d_(1)` and external diameter `d_(2)`, where `d_(2)=3d_(1)`. On heating , `d_(2)` increases by `0.3%`. Then `d_(1)` will :-A. decreases by ` 0.1 %`B. decrease by `0.3%`C. increases by `0.1%`D. increase by `0.3%`.

A faulty thermometer reads freezing point and boiling point of water as `-5 ^(@)C` abd `95^(@)C` respectively . What is the correct value of temperature as it reads `60^(@)C` on faulty thermometer?A. `60^(@)C`B. `65^(@)C`C. `64^(@)C`D. `62^(@)C`

A convex lens of focal length 20cm and a concave lens of focal length f are mounted coaxially 5cm apart. Parallel beam of light incident on the convex lens emerges from the concave lens as a parallel beam. Then, f in cm isA. 35B. 25C. 20D. 15

A convex lens A of focal length 20cm and a concave lens G of focal length 5cm are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance d in cm will beA. 25B. 15C. 30D. 50

A convex lens forms an image of an object placed 20cm away from it at a distance of 20cm on the other side of the lens. If the object is moves 5cm toward the lens, the image will beA. `5 cm` towards the lensB. `5 cm` away from the lensC. `10 cm` towards the lensD. `10 cm` away from the lens

The graph shows how the magnification `m` produced by a convex thin lens varies with image distance `v`. What was the focal length of the used ? A. `(b)/(c)`B. `(b)/(ca)`C. `(bc)/(a)`D. `(c)/(b)`

A convergent beam of light converges to a point `20 cm` behind the convex mirrorr on the principal axis.An inverted image of the same size is formed coincident with the virtual object. Then, the focal length of the convex mirrorr isA. `20 cm`B. `10 cm`C. `40 cm`D. `30 cm`

A Galileo telescope has an objective of focal length `100cm` and magnifying power 50. The distance between the two lenses in normal adjustment will beA. `98 cm`B. `100cm`C. `150cm`D. `200 cm`

A glass `(mu = 1.5)` sphere of radius R is viewed from outside along a diameter. Calculate the distance between two points (say P and Q) lying on the line AB whose images are seen at centre C and point A respectively.

A large rectangular glass block of refractive index `mu` is lying on a horizontal surface as shown in Figure. Find the minimum value of `mu` so that the spot S on the surface cannot be seen through top plane ‘ABCD’ of the block.

The focal length of the lens used in question 118 isA. `(D^(2)+d^(2))/(2D)`B. `(D^(2)-d^(2))/(4D)`C. `(D^(2)-d^(2))/(2D)`D. `(D^(2)+d^(2))/(d)`

Can we project the image formed by a plane mirror on to a screen? Give reasons.

A ray of light travelling in the direction `(1)/(2)``(hati,+sqrt3hatj)` is incident on a plane mirror. After reflection, it travels along the direction (1)/(2)` `(hati-sqrt3hatj)` . The angle of incidence isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `75^(@)`

A ray of light in incident along a vector `hati + hatj - hatk` on a plane mirror lying in y-z plane. The unit vector along the reflected ray can beA. `(hati + hatj -hatk)/(sqrt(3))`B. `(hati - hatj +hatk)/(sqrt(3))`C. `(-hati + hatj -hatk)/(sqrt(3))`D. `(3hati + hatj -hatk)/(sqrt(3))`

A ray reflected successively from two plane mirrors inclined at a certain angle `(lt 90^(@))` undergoes a deviation of `300^(@)` . The number of images observable are:A. 10B. 11C. 12D. 14

An air bubble in a glass sphere `(mu = 1.5)` is situated at a distance `3 cm` from a convex surface of diameter `10 cm`. At what distance from the surface will the bubble appear ?