804 + Interview Questions in OPTICS IN GENERAL AWARENESS Page 8 InterviewSolution

351.	Explain the working of an endoscope.
Answer» An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient’s body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body. Even operations could be carried out with the endoscope cable which has the necessary instruments attached at their ends.

Discussion

352.	A train of plane light waves propagates in the medium where the phase velocity `v` is a linear function of wavelength: `v = 1 + blambda`, where `a` and `b` are some positive constants. Demonstrate that in such a medium the shape of an arbitary train of light waves is restroed after the time interval `tau = 1//b`.
Answer» We write `v = (omega)/(k)= a+ b lambda` so `omega = k(a+b lambda) = 2pi b + ak`. (since `k = (2pi)/(lambda)`). Suppose a wavetrain at time `t = 0` has the form `F(x, 0) = int f(k) e^(ikx) dk` Then at time `t` it will have the form `F(x,t) = int f(k) e^(ikx-i omegat) dk` `=int f(k)e^(ikx -i(2pib + ak)t) = int f(k)e^(ik(x-at)e^(-2pi bt) dk` At `t = (1)/(b) = tau` `F(x, tau) = F(x - a tau, 0)` so at time `t = tau` the wave train has regained its shape through it has advanced by `a tau`.

Discussion

353.	A student uses spectacles of number -2 for seeing distant objects. Commonly used lenses for her/his spectacles are (A) bi-concave (B) piano concave (C) concavo-convex (D) convexo-concave
Answer» Correct answer is (A) bi-concave

Discussion

354.	Obtain the equation for critical angle.
Answer» Snell’s law in the product form, equation for critical angle incidence becomes, n₁ sini i_c = n₂ sin 90° n₁ sini i_c = n₂ (∵ sin 90° = 1) sini i_c = (\(\frac{n_2}{n_1}\)) Here, n₁ > n₂ If the rarer medium is air, then its refractive index is 1 and can be taken as n itself, i.e. (n₂ = 1) and (n₁ = n). sini i_c = \(\frac{1}{n}\) or i_c= sin^-1(\(\frac{1}{n}\))

354.

Obtain the equation for critical angle.

Answer»

Snell’s law in the product form, equation for critical angle incidence becomes, n₁ sini i_c = n₂ sin 90°

n₁ sini i_c = n₂ (∵ sin 90° = 1)

sini i_c = (\(\frac{n_2}{n_1}\))

Here, n₁ > n₂

If the rarer medium is air, then its refractive index is 1 and can be taken as n itself, i.e. (n₂ = 1) and (n₁ = n).

sini i_c = \(\frac{1}{n}\) or i_c= sin^-1(\(\frac{1}{n}\))

Discussion

355.	Mirage is accounted for by………….. .
Answer» total internal reflection.

Discussion

356.	Explain the reason for glittering of diamond.
Answer» Diamond appears dazzling because the total internal reflection of light happens inside the diamond. The refractive index of only diamond is about 2.417. It is much larger than that for ordinary glass which is about only 1.5. The critical angle of diamond is about 24.4°. It is much less than that of glass. A skilled diamond cutter makes use of this larger range of angle of incidence (24.4° to 90° inside the diamond), to ensure that light entering the diamond is total internally reflected from the many cut faces before getting out. This gives a sparkling effect for diamond.

Discussion

357.	For a wavelength of light ‘λ’ and scattering object of size ‘a’ , all wavelength are scattered nearly equally, if(a) a = λ (b) a >> λ (c) a << λ (d) a ≥ λ
Answer» (b) a >> λ For a >> λ, the scattering power is not selective.

357.

For a wavelength of light ‘λ’ and scattering object of size ‘a’ , all wavelength are scattered nearly equally, if(a) a = λ (b) a >> λ (c) a << λ (d) a ≥ λ

Answer»

(b) a >> λ

For a >> λ, the scattering power is not selective.

Discussion

358.	In optical fibres, the refractive index of the core is (a) greater than that of the cladding (b) equal to that of the cladding (c) smaller than that of the cladding (d) independent of that of the cladding
Answer» (a) greater than that of the cladding In optical fibres, refractive index of core material > refractive index of the cladding.

358.

In optical fibres, the refractive index of the core is (a) greater than that of the cladding (b) equal to that of the cladding (c) smaller than that of the cladding (d) independent of that of the cladding

Answer»

(a) greater than that of the cladding

In optical fibres, refractive index of core material > refractive index of the cladding.

Discussion

359.	Consider following phenomena/applications : P) Mirage, Q) rainbow, R) Optical fibre and S) glittering of a diamond. Total internal reflection is involved in(A) Only R and S(B) Only R (C) Only P, R and S (D) all the four
Answer» Correct answer is (A) Only R and S

Discussion

360.	The `6563 Å H_(2)` line emitted by hydrogen in a star is found to be red shifted by `15 Å`. Estimate the speed with which the star is receding from earth.
Answer» Here, `lambda = 6563 Å, Delta lamnda = + 15 Å, c = 3 xx 10^(8)ms^(-1)` Since the star is receding away, hence its velocity `v` is negative (i.e. if `Delta lambda` positive, `v` is negative) `:. Delta lambda = - v lambda//c or v = -(c Delta lambda)/(lambda) = -(3 xx 10^(8) xx 15)/(6563) = - 6.86 xx 10^(5) ms^(-1)` Here, negative sign shows recession of star.

Discussion

361.	What are the two types of spherical mirrors ?
Answer» Concave spherical mirror and convex spherical mirror.

Discussion

362.	Which spherical mirror is converging and which one is diverging ?
Answer» A concave spherical mirror is converging and a convex spherical mirror is diverging.

Discussion

363.	Which spherical mirror forms a virtual, erect and smaller image of an object ?
Answer» A convex spherical mirror forms a virtual, erect and smaller image of an object.

Discussion

364.	Which spherical mirror has a real focus and which one has a virtual focus ?
Answer» A concave spherical mirror has a real focus and a convex spherical mirror has a virtual focus.

Discussion

365.	An object of size `10 cm` is placed at a distance of `50 cm` from a concave mirror of focal length `15 cm`. Calculate location, size and nature of the image.
Answer» Correct Answer - `-21.4 cm`, `-4.28 cm` ; Real

Discussion

366.	How does focal length of a convex lens change if violet light is used instead of red light.
Answer» `(1)/(f_(v)) = (mu_(v) - 1)((1)/(R_(1)) - (1)/(R_(2)))` `(1)/(f_(r )) = (mu_(r ) - 1)((1)/(R_(1)) - (1)/(R_(2)))` As `mu_(v) gt mu_(r ) :. f_(v) lt f_(r )` i.e., focal length decreases.

Discussion

367.	position of the object place infront of a convex lens are given in column I match them with the natures of the images formed by the convex lens given in column II.Column IColumn IIABeyond centre of curvature(i)real image at infinityBBetween F and C(ii)virtual and larger imageCAt the principal focus(iii)virtual at infinityDBetween F and optical centre(iv)same as the object
Answer» A. (v) B. (iv) C. (i) D. (ii)

367.

position of the object place infront of a convex lens are given in column I match them with the natures of the images formed by the convex lens given in column II.Column IColumn IIABeyond centre of curvature(i)real image at infinityBBetween F and C(ii)virtual and larger imageCAt the principal focus(iii)virtual at infinityDBetween F and optical centre(iv)same as the object

Answer»

A. (v)

B. (iv)

C. (i)

D. (ii)

Discussion

368.	What is focal length and power of a rectangular glass slab ?
Answer» Focal length of a glass slab `= oo` and power of glass slab `= Zero`

Discussion

369.	A lens of focal length f is placed in between an object and screen at a distance D. The lens forms two real images of object on the screen for two of its differenct positions, a distance x apart. The two real images have magnifications `m_(1)` and `m_(2)`, respectivelly `(m_(1)gtm_(2))`. Then,A. `d gt 2f`B. `d gt 4gf`C. `M_(1)M_(2)=1`D. `\|M_(1)-\|M_(2)\|=1`
Answer» Correct Answer - B::C

Discussion

370.	A concave mirror produces real image `10 mm` tall, of an object `2.5 mm` tall placed at `5 cm` from the mirror. Calculate focal length of the mirror and the position of the image.
Answer» Correct Answer - `- 4 cm ; - 20 cm`

Discussion

371.	What is the refractive index of a medium in which light travels with a speed of `2 xx 10^(8)m//s` ?A. `3//2`B. `2//3`C. `1`D. none of these
Answer» Correct Answer - A

Discussion

372.	The image formed by a convex mirror of focal length `30 cm`. is a quarter of the object. What is the distance of the object from the mirror ?
Answer» Correct Answer - `- 90 cm`

Discussion

373.	An object is placed `0.4 m` from a convex mirror and a plane mirror is placed at a distance of `0.3 m` from the object. The images formed in the two mirrors coincide without parallex. What is the focal length of the convex mirror ?
Answer» Correct Answer - `0.4 m`

Discussion

374.	An object of size `3.0 cm` is placed `14 cm` in front of a concave lens of focal length `21 cm`. Describe the image produced by the lens. What happens if the object is moved further from the lens ?
Answer» Here, `h_(1) 3cm , u = -14 cm, f = - 21 cm, v = ?` As `(1)/(v) - (1)/(u) = (1)/(f)` `(1)/(v) = (1)/(f) + (1)/(u) = (1)/(-21) + (1)/(-14) = (-2 - 3)/(42) = (-5)/(42)` `v = (-42)/(5) = -8.4 cm` `:.` Image is erect, virtual and at `8.4 cm` from the lens on the same side as the object. As `(h_(2))/(h_(1)) = (v)/(u) :. (h_(2))/(3) = (-8.4)/(-14)` `h_(2) = 0.6 xx 3 = 1.8 cm` As the object is moved away from the lens, virtual image moves towards focus of lens (but never beyond focus). The size of image goes on decreasing.

Discussion

375.	A concave mirror of focal length `20 cm` is placed at a distance of `50 cm` from a wall. How far from the wall should an object be placed to form its real image on the wall ?
Answer» Correct Answer - `16.7 cm`

Discussion

376.	A concave mirror of focal length `10 cm` is placed at a distance of `35 cm` from a wall. How far from the wall should an object be placed to get its image on the wall ?
Answer» Correct Answer - `21 cm` Here, `f = - 10 cm, v = - 35 cm`, `u = - 14 cm` `:.` Distance of object from the wall `= 35 -14 = 21 cm`

Discussion

377.	What kind of polarizations has a plane electromagetic wave if the projections of the vector `E` on the `x` and `y` axes are perpendicular to the propagation direction and are defind by the following equations: (a) `E_(x) = Ecos (omegat - kz), E_(y) = E sin (omegat - kz)`, (b) `E_(x) = E cos(omegat - kz), E_(y) = E cos (omegat - kz + pi//4)` (c) `E_(x) = E cos (omega t - kz), E_(y) = E cos (omega t - kz + pi)`?
Answer» The wave is moving in the direction of `z`-axis (a) Here `E_(x) = E cos (omegat - kz), E_(y) = E sin(omegat - kz)` `(E_(x)^(2))/(E^(2)) + (E_(y)^(2))/(E^(2)) = 1` so the up of the electric vector moves along a circle. For the right handed coordinate system this represents circular anticlockwise polarization when observed towards the incoming wave. (b) `E_(x) = E cos (omegat - kz), E_(y) = Ecos (omegat - kz+(pi)/(4))` so `(E_(y))/(E) = (1)/(sqrt(2)) co s(omegat - kz) - (1)/(sqrt(2)) sin (omegat - kz)` or `((E_(y))/(E) - (1)/(sqrt(2))(E_(x))/(E))^(2) = (1)/(2)(1-(E_(x)^(2))/(E^(2)))` or `(E_(y)^(2))/(E^(2)) + (E_(x)^(2))/(E^(2)) - sqrt(2) (E_(y)E_(x))/(E^(2)) = (1)/(2)` This is clearly an ellipse. By comparing with the pervious case (compare the phase of `E_(y)` in the two cases ) we see this represents elliptical clockwise polarization when viewed towards the incoming wave. We write the equations as `E_(x) + E_(y) = 2Ecos(omegat - kz+(pi)/(8)) cos((pi)/(8))` `E_(x) - E_(y) = +2E sin (omegat - kz + (pi)/(8)) sin ((pi)/(8))` Thus `((E_(x) - E_(y))/(2Ecos((pi)/(8))))^(2) + ((E_(x) - E_(y))/(2esin((pi)/(8))))^(2) = 1` Since `cos((pi)/(8)) gt sin((pi)/(8))`, the major axis is the direction of the stright line `y = x`. (c) `E_(x) = Ecos (omegat - kz)` `E_(y) = Ecos(omegat - kz + pi) =- E cos (omegat - kz)` Thus the top of the electric vector traces the curve `E_(y) =- E_(x)` which is a stright line `~(y =- x)`. It corresponds to plane polarization.

Discussion

378.	An object is held in front of conacve mirror of focal length `15 cm`. The image formed is `3` times the size of the object. Calculate two possible distances of the object from the mirror.
Answer» Correct Answer - `-10 cm`; `-20 cm` Here, `f = - 15 cm, m = +- 3, u = ?` As `m = -(v)/(u) = + 3, v = - 3 u` From `(1)/(v) + (1)/(u) = (1)/(f)`, `(1)/(-3u) + (1)/(u) = (1)/(-15)` or `(2)/(3 u) = (1)/(-15)`, `u = -10 cm` Similarly, when `m = -(v)/(u) = - 3`, we get `u = - 20 cm`

Discussion

379.	The speed of light in media `M_1` and `M_2` are `1.5 xx 10^8 m//s and 2.0 xx 10^8 m//s` respectively. A ray of light enters from medium `M_1` to `M_2` at an incidence angle `i`. If the ray suffers total internal reflection, the value of `i` is.A. Equal to `sin^(-1)((2)/(3))`B. Equal to or less than `sin^(-1) ((3)/(5))`C. Equal to or greater than `sin^(-1)((3)/(4))`D. Less than `sin^(-1) ((2)/(3))`
Answer» Correct Answer - C Given , `v_(1) = 1.5 xx 10^(8) m//s , v_(2) = 2.0 xx 10^(8) m//s` Refractive index for medium `M_(1)` is `mu_(1) = ( c)/(v_(1)) = (3 xx 10^(8))/(1.5 xx 10^(8)) = 2` Refractive index for medium `M_(2)` is `mu_(2) = ( c)/(v_(2)) = (3 xx 10^(8))/(2.0 xx 10^(8)) = (3)/(2)` If `i` is the angle of incidence and `C` is the critical angle then for total internal reflection `sin i ge sin C` But `sin C = mu_(2)//mu_(1)` `:. sin i ge (mu_(2))/(mu_(1)) ge (3//2)/(2) or i ge sin^(-1)((3)/(4))`

Discussion

380.	An object of height `h` is held before a spherical mirror of focal length `\| f \| = 40 cm`. The image of the object produced by the mirror has same orientation as the object and has `height = 0.2 h`. Is the image real or virtual ? Is the image on the convex or concave? What is focal length of mirror with proper sign?
Answer» As the image of the object has the same orientation as the object, the image must be virtual, and on the opposite side of the object. Now, `m = (h_(2))/(h_(1)) = -(v)/(u) = 0.2` `:. v = -0.2 u` From `(1)/(f) = (1)/(v) + (1)/(u)` , `(1)/(f) = (1)/(-0.2 u) + (1)/(u) = -(4)/(u)` `f = -(u)/(4)`. As `u` is negative, `f` must be positive, i.e., `f = + 40 cm`. The mirror must be convex.

Discussion

381.	Prove that spherical mirror formula is applicable equally to a plane mirror.
Answer» The spherical mirror formula is `(1)/(u) + (1)/(v) = (1)/(f)` …(i) For a plane mirror, `R = oo` `f = (R )/(2) = oo` Form (i), `(1)/(u) + (1)/(v) = (1)/(oo) = 0` or `(1)/(v) = -(1)/(u)` or `v = -u` As `u` is negative, `v` becomes +. Hence, image is formed behind the mirror at the same distance as the object is in front of it. This is what happens in a plane mirror. Hence the desired result.

Discussion

382.	The relation governing refraction of light from rarer to denser medium at a spherical refracting surface isA. `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`B. `(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`C. `(mu_(1))/(u) - (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`D. none of these
Answer» Correct Answer - A

Discussion

383.	The wall of a room is covered with a perfect plane mirror and two movie films are made, one recording the movement of a man and the other of his mirror image. While viewing the film later, can an outside tell which is which ?
Answer» Nature exhibits left right symmetric i.e. physical laws are the same for an object and its mirror image. Therefore, an outsider cannot distinguish between the two films. ltbr? However, the distinguish can be made if there is any initial asymmetric information.

Discussion

384.	A transparent diffraction grating has a period `d = 1.50mum`. Find the angular disperison `D` (in angular minutes per nanometers) corresponding to the maximum of highest order for a spectral line of wavelength `lambda = 530mm` of light falling on the grating (a) at right angles, (b) at the angle `theta_(0) = 45^(@)` to the normal.
Answer» For normal incidence, the maxima are given by `d sin theta= n lambda` so `sintheta = n(lambda)/(d) = nxx (0.530)/(1.500)` Clearly `nle 2` as `sin theta gt 1` for `n = 3`. Thus the highest order is `n = 2`. Then `D=(d theta)/(d lambda) = (k)/(d cos theta) = (k)/(d) (1)/(sqrt(1-((k lambda)/(d))^(2)))` Putting `k =2, lambda =0.53mum, d= 1.5mum = 1500 nm` we get `D = (2)/(1500) (1)/(sqrt(1-((1.06)/(1.5))^(2))) xx (180)/(pi) xx 60 =6.47ang.min//nm`. (b) We write the diffraction formula as `d(sin theta_(0) + sin theta) = k lambda` so `sin theta_(0) + sin theta = k (lambda)/(d)` Here `theta_(0) = 45^(@)` and `sin theta_(0) = 0.707` so `sin theta_(0)+sintheta le 1.707`. Since `(lambda)/(d) = (0.53)/(1.5) = 0.353333`, we see that `k le4` Thus highest order corresponding to `k = 4`. Now as before `D = (d theta)/(d lambda)`so `D = (k)/(d cos theta) = (k//d)/(sqrt(1-((k lambda)/(d)-sintheta_(0))^(2)))` `= 12.948ang. min//nm`,

Discussion

385.	The formula governing reflection of light from a spherical mirror is `(1)/(v) + (1)/(u) = (2)/(R )`, where `u` = distance of object from pole of mirror, `upsilon` = distance of image from pole of mirror `f` = focal length of mirror, `R` = radius of curvature of mirror. This is known as mirror formula and is applicable equally to concave mirror and convex mirror. `m = (I)/(O) = (upsilon)/(u)` Read the above passage and answer the following questions : (i) An object is held at a distance of `30 cm` in front of a concave mirror of radius of curvature `40 cm`. Calculate distance of the image from the object ? What is linear magnification of the mirror ? (ii) The object is moved to a distance of `40 cm` in front of the mirror. How is focal length of mirror affected ? (iii) What values of life do you learn from the mirror formula ?
Answer» (i) Here, `u = -30 cm, R = - 40 cm, v = ?` From `(1)/(v) + (1)/(u) = (1)/(f) = (2)/(R )` `(1)/(v) = (2)/( R) - (1)/(u) = (2)/(-40) + (1)/(30) = -(1)/(60)` `v = - 60cm`, on the same side as object. `:.` Distance of image from the object `= 60 - 30 = 30 cm`. `m = (I)/(O) = (-v)/(u) = (+60)/(-30) = -2`, negative sign for inverted image. (ii) Focal length `(f)` of mirror remains unaffected. On changing `u` , `v` changes and not `f`. (iii) Mirror formula reveals that `f` depends only on `R`, and not `u` or `v`. Infact on changing `u` , `v` changes, but `f` remains constant. In day to day life, `u` corresponds to a situation that arises and `v` corresponds to our response to the situation. We are like a mirror. Our nature/curvature determines our focal length. The mirror formula implies that our nature is not affected by the situation that comes up. Response to a particular situation will depend on our nature.

Discussion

386.	In single slit diffraction experiment, yellow light is replaced by X-rays. How will the diffraction pattern be affected ?
Answer» The diffraction pattern will disappear as wavelength of X-rays is much smaller than the wavelength of yellow light. The essential condition for diffraction, i.e., `a ~~ lambda` is no longer satisfied.

Discussion

387.	The rays of light falling on a convex lens in a direction parallel to principal axis of the lens, get refracted through the lens and meet actually at a single point `F` on the principal axis of the lens. This point is called principal focus of the lens. Read the above passage and answer the following questions : (i) Is principal focus of a convex lens, a real point ? Is the same true for a concave lens ? (ii) A distinct image of a distant tree os obtained on a screen held at `40 cm` from a convex lens. What is its focal length ? (iii) Our teachers and parants advise us to stay focussed. What does it imply ?
Answer» (i) Yes, the principal focus of a convex lens is a real point. This is because rays refracted through convex lens meet actually at this point. No, principal focus of a concave lens is a virtual point only. (ii) f = distance of screen from the lens = 40 cm (iii) Our parants and teachers advise us to stay focussed. It implies that we concentrate all our energies/efforts at a single point/problem so that we can resolve the same easily. Staying focussed means that we do not divert our energies and attention to several things at a time. This would lead us nowhere. Thus, the secret of success is to stay focussed.

Discussion

388.	Light with wavelength `lambda` falls on a diffracting grating at right angles. Find the angular dispersion of the grating as a function of diffraction angle `theta`.
Answer» We have `d sin theta= k lambda` so `(d theta)/(d lambda) = D = (k)/(d cos theta) = (tan theta)/(lambda)`

Discussion

389.	A swimmer S inside water is vertically above a fixed point P. A rectangular glass slab B is placed between S and P. As seen by S, the position of P will appear to change, if (a) B is moved horizontally (b) B is moved vertically (c) S moves horizontally (d) S moves vertically
Answer» Correct Answer is: (c) S moves horizontally For normal viewing, apparent change in the position of P due to the slab does not depend on the position of the slab. When S moves horizontally, the condition of normal viewing does not apply, and the apparent position of P will change with the angle of viewing.

389.

A swimmer S inside water is vertically above a fixed point P. A rectangular glass slab B is placed between S and P. As seen by S, the position of P will appear to change, if (a) B is moved horizontally (b) B is moved vertically (c) S moves horizontally (d) S moves vertically

Answer»

Correct Answer is: (c) S moves horizontally

For normal viewing, apparent change in the position of P due to the slab does not depend on the position of the slab. When S moves horizontally, the condition of normal viewing does not apply, and the apparent position of P will change with the angle of viewing.

Discussion

390.	An intially parallel cyclindrical beam travels in a medium of refractive index `mu (I) = mu_(0) + mu_(2) I`, where `mu_(0)` and `mu_(2)` are positive constants and `I` is intensity of light beam. The intensity of the beam is decreasing with increasing radius. Answer the following questions : The initial shape of the wavefront of the beam isA. concaveB. convex near the axis and concave near the peripheryC. planarD. convex
Answer» Correct Answer - C The initial shape of the wave front of the beam is planar, as the wave front lies in the plane of the beam.

Discussion

391.	A bird flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the bird will appear to (a) be farther away than its actual distance (b) be closer than its actual distance (c) move faster than its actual speed (d) move slower than its actual speed
Answer» Correct Answer is: (a, c) Let x = height of the bird above the water surface. For light travelling from the bird to the fish, μ₁ = 1, μ₂ = μ (refractive index of water), u = -x. μ/v = 1/-x or v = -μx or \|v\| = μx > x. Speed of bird = x , apparent speed of bird = \|v\| = μx.

391.

A bird flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the bird will appear to (a) be farther away than its actual distance (b) be closer than its actual distance (c) move faster than its actual speed (d) move slower than its actual speed

Answer»

Correct Answer is: (a, c)

Let x = height of the bird above the water surface.

For light travelling from the bird to the fish,

μ₁ = 1, μ₂ = μ (refractive index of water), u = -x.

μ/v = 1/-x

or v = -μx

or |v| = μx > x.

Speed of bird = x , apparent speed of bird = |v| = μx.

Discussion

392.	A brid flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the bird will appear toA. be farther away than its actual distanceB. be closer than its actual distanceC. move faster than its actual speedD. move slower than its actual speed
Answer» Correct Answer - A::C

Discussion

393.	A plane wavefront is incident on a plane reflecting surface at an angle of `30^(@)`. What angle will the reflected rays make with the reflecting surface ?
Answer» Here, `I = 30^(@)` According to the laws of reflection, `r = I = 30^(@)` :. the reflected wavefront will make an angle of `30^(@)` with the plane surface. The refracted rays will make an angle of `(90^(@) + 30^(@)) = 120^(@)` with the plane surface.

Discussion

394.	What is the shape of interference fringes in YDSE ?
Answer» The fringes are nearly sraight line fringes parallel to the slits.

Discussion

395.	what is the relation between path diff. and wavelength for destructive interference ?
Answer» Path diff. `=(2n - 1)lambda//2`, where `n = 1, 2, 3,…..` i.e., path diff. should be an odd integral multiple of half the wavelength of light.

Discussion

396.	Statement-1 : A ray incident along normal to the mirror retraces its path. Statement-2 : In reflection, angle of incidence is always equal to angle of reflectionA. If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I.B. If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I.C. If Statement - I is true but Statement - II is false.D. If Statement - I is false but Statement - II is true.
Answer» Correct Answer - A

Discussion

397.	A ray incident along normal to the mirror retraces its path. Why ?
Answer» This is because `anglei = angle r = 0^(@)`.

Discussion

398.	A body of height `1m` stands in front of a convex mirro. His distance from the mirror is equal to its focal length. The height of his image isA. `0.25m`B. `0.33m`C. `0.5m`D. `0.67m`
Answer» Correct Answer - C

Discussion

399.	What is the value of focal length of a plane mirror ?
Answer» Focal length of a plane mirror is infinity. Power `P = (1)/(f) = (1)/(oo) = Zero`

Discussion

400.	An object is placed at a distance of 0.5 m infront of a plane mirror. The distance between object and image will be(a) 0.25 m(b) 0.5 m (c) 1.0 m (d) 2.0 m
Answer» (c) 1.0 m Distance between object and image = 0.5 + 0.5 = 1.0 m.

400.

An object is placed at a distance of 0.5 m infront of a plane mirror. The distance between object and image will be(a) 0.25 m(b) 0.5 m (c) 1.0 m (d) 2.0 m

Answer»

(c) 1.0 m

Distance between object and image = 0.5 + 0.5 = 1.0 m.

Discussion

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Explain the working of an endoscope.

A student uses spectacles of number -2 for seeing distant objects. Commonly used lenses for her/his spectacles are (A) bi-concave (B) piano concave (C) concavo-convex (D) convexo-concave

Obtain the equation for critical angle.

Mirage is accounted for by………….. .

Explain the reason for glittering of diamond.

For a wavelength of light ‘λ’ and scattering object of size ‘a’ , all wavelength are scattered nearly equally, if(a) a = λ (b) a &gt;&gt; λ (c) a &lt;&lt; λ (d) a ≥ λ

In optical fibres, the refractive index of the core is (a) greater than that of the cladding (b) equal to that of the cladding (c) smaller than that of the cladding (d) independent of that of the cladding

Consider following phenomena/applications : P) Mirage, Q) rainbow, R) Optical fibre and S) glittering of a diamond. Total internal reflection is involved in(A) Only R and S(B) Only R (C) Only P, R and S (D) all the four

The `6563 Å H_(2)` line emitted by hydrogen in a star is found to be red shifted by `15 Å`. Estimate the speed with which the star is receding from earth.

What are the two types of spherical mirrors ?

Which spherical mirror is converging and which one is diverging ?

Which spherical mirror forms a virtual, erect and smaller image of an object ?

Which spherical mirror has a real focus and which one has a virtual focus ?

An object of size `10 cm` is placed at a distance of `50 cm` from a concave mirror of focal length `15 cm`. Calculate location, size and nature of the image.

How does focal length of a convex lens change if violet light is used instead of red light.

What is focal length and power of a rectangular glass slab ?

A concave mirror produces real image `10 mm` tall, of an object `2.5 mm` tall placed at `5 cm` from the mirror. Calculate focal length of the mirror and the position of the image.

What is the refractive index of a medium in which light travels with a speed of `2 xx 10^(8)m//s` ?A. `3//2`B. `2//3`C. `1`D. none of these

The image formed by a convex mirror of focal length `30 cm`. is a quarter of the object. What is the distance of the object from the mirror ?

An object is placed `0.4 m` from a convex mirror and a plane mirror is placed at a distance of `0.3 m` from the object. The images formed in the two mirrors coincide without parallex. What is the focal length of the convex mirror ?

An object of size `3.0 cm` is placed `14 cm` in front of a concave lens of focal length `21 cm`. Describe the image produced by the lens. What happens if the object is moved further from the lens ?

A concave mirror of focal length `20 cm` is placed at a distance of `50 cm` from a wall. How far from the wall should an object be placed to form its real image on the wall ?

A concave mirror of focal length `10 cm` is placed at a distance of `35 cm` from a wall. How far from the wall should an object be placed to get its image on the wall ?

An object is held in front of conacve mirror of focal length `15 cm`. The image formed is `3` times the size of the object. Calculate two possible distances of the object from the mirror.

Prove that spherical mirror formula is applicable equally to a plane mirror.

The relation governing refraction of light from rarer to denser medium at a spherical refracting surface isA. `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`B. `(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`C. `(mu_(1))/(u) - (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`D. none of these

The wall of a room is covered with a perfect plane mirror and two movie films are made, one recording the movement of a man and the other of his mirror image. While viewing the film later, can an outside tell which is which ?

In single slit diffraction experiment, yellow light is replaced by X-rays. How will the diffraction pattern be affected ?

Light with wavelength `lambda` falls on a diffracting grating at right angles. Find the angular dispersion of the grating as a function of diffraction angle `theta`.

A swimmer S inside water is vertically above a fixed point P. A rectangular glass slab B is placed between S and P. As seen by S, the position of P will appear to change, if (a) B is moved horizontally (b) B is moved vertically (c) S moves horizontally (d) S moves vertically

A bird flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the bird will appear to (a) be farther away than its actual distance (b) be closer than its actual distance (c) move faster than its actual speed (d) move slower than its actual speed

A brid flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the bird will appear toA. be farther away than its actual distanceB. be closer than its actual distanceC. move faster than its actual speedD. move slower than its actual speed

A plane wavefront is incident on a plane reflecting surface at an angle of `30^(@)`. What angle will the reflected rays make with the reflecting surface ?

What is the shape of interference fringes in YDSE ?

what is the relation between path diff. and wavelength for destructive interference ?

A ray incident along normal to the mirror retraces its path. Why ?

A body of height `1m` stands in front of a convex mirro. His distance from the mirror is equal to its focal length. The height of his image isA. `0.25m`B. `0.33m`C. `0.5m`D. `0.67m`

What is the value of focal length of a plane mirror ?

An object is placed at a distance of 0.5 m infront of a plane mirror. The distance between object and image will be(a) 0.25 m(b) 0.5 m (c) 1.0 m (d) 2.0 m

For a wavelength of light ‘λ’ and scattering object of size ‘a’ , all wavelength are scattered nearly equally, if(a) a = λ (b) a >> λ (c) a << λ (d) a ≥ λ