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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
A small object (A) is placed on the principal axis of an equiconvex lens at a distance of 30 cm. The refractive index of the glass of the lens is 1.5 and its surfaces have radius of curvature `R = 20 cm`. Two glass slabs S1 and S2 have been placed behind the lens as shown in Figure. Thickness of the two slabs is 6 cm and 4 cm respectively and their refractive indices are `(3)/(2)` and 2 respectively. (a) Find the distance of the final image measured from the lens. Also find the magnification. (b) How does the position of the image change if the slab S2 is moved to left so as to put it in contact with S1. (c) How does the position of the image change if the two slabs in contact are together moved to right by a distance of 100 cm. |
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Answer» Correct Answer - (a) `60 cm; m =- 2` (b) No change (c) No change |
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| 552. |
A small telescope has an objective lens of focal length `144 cm` and an eye-piece of focal length `6.0 cm`. What is the magnifying power of the telescope ? What is the separation between the objective and the eye-piece ? |
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Answer» Here, `f_(0) = 144 "cm" , f_(e) = 6.0"cm" , MP = ? L =?` `MP = (-f_(0))/(f_(e)) = (-144)/(6.0) = -24 "and" L = f_(0) + f_(e) = 144 + 6.0 = 150.0 "cm"` |
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| 553. |
There is a spherical glass shell of refractive index 1.5, inner radius 10 cm and outer radius 20 cm. Inside th espherical cavity, there is air. A point object is placed at a point O at a distance of 30cm from the outer spherical surface. Find the final position of th eimage as seen eye. |
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Answer» Correct Answer - `18.75` cm from centre. `(1.5)/(v)-(1)/(-30)=(1.5-1)/(20)rArrv=-180cm` For second surface, `u_(1)=-190cm` `rArr (1)/(v)-(1.5)/(-190)=(1-1.5)/(10)rArr v=(190)/(11)` For third surface, `u=-((190)/(11)+20)=(-410)/(11)` `rArr(1.5)/(v)-(11)/(410)=(1.5-1)/(-10)=-(1)/(20)` `rArr v=-(410)/(21)` For fourth surface , `u=-((410)/(21)+10)=-(620)/(21)cm` `mu_(1)=1.5, mu_(2)=1,R=-20cm` `rArr (1)/(v)+(1.5xx21)/(620)=(1-1.5)/(-20)` `rArr v=-38.75 cm` The final image is at a distance of `18.75cm` from the center. |
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| 554. |
A far sighted person has a near point of `60"cm"` . What power lens should be used for eye glasses such that the person can be read this book at a distance of `25` cm. |
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Answer» Here `v=-60"cm" , u=-25"cm"` `(1)/(f) = (1)/(v) - (1)/(u) = -(1)/(60) + (1)/(25) implies f= (300)/(7) "cm" therefore "Power" = (1)/("f(inm") = (1)/((3//7)) = +2.33D` |
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| 555. |
The layered lens shown in Figure. , is made of kinds of glasses. How many and what kind of images will be produced by this lens with a point source placed on the optical axis? Neglect the reflection of light at the boundaries between layers. |
| Answer» Two images will be formes, one from each type of material. | |
| 556. |
A thin glass (refractive index 1.5) lens has optical power of `-5D` in air. Its optical power in a liquid medium with refractive index 1.6 will beA. 1DB. `-1D`C. `25 D`D. `-25D` |
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Answer» Correct Answer - A Power of lens = `mu_(2) - mu_(1) ((1)/(R_(t) - (1)/(R_(2)))` in air power is -5D `therefore -5 = (1.5 -1) ((1)/(R_(1)) - (1)/(R_(2)))` ……. (i) In other medium power is P `P = (1.5 - 1.6) ((1)/(R_(1))- (1)/(R_(2)))` ……. (ii) Dividing (ii) by (i) `(P)/(-5) = (-0.1)/(1) xx (1)/(0.5) implies P = 1D` |
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| 557. |
A thin glass (refractive index 1.5) lens has optical power of `-5D` in air. Its optical power in a liquid medium with refractive index 1.6 will beA. `25 D`B. `-25 D`C. `1 D`D. None of these |
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Answer» Correct Answer - D `(f_(1))/(f_(a))=((._(a)mu_(g)-1))/((._(a)mu_(g)-1))` `implies(f_(l))/(f_(a))=(._(a)mu_(g)-1)/(._(a)mu_(g)-1)=(1.5-1)/((1.5)/(1.6)-1)=(0.5xx1.6)/(-0.1)=-8` `impliesP_(l)=(P_(a))/(8)=(5)/(8)` |
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| 558. |
While looking at her face in a mirror, Hema notes that her face is highly magnified when she is close to the mirror. As she backs away from the mirror, her image first gets blurry, then disappears when she is at a distance of 45 cm from the mirror. Explain the happenings? What will happen if she moves beyond 45 cm distance from the mirror? |
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Answer» Correct Answer - Image gets inverted beyond 45 cm |
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| 559. |
Far points of myopic eye is `250 cm,` then the focal length of the lens to be used will beA. `-250 cm`B. `-250//9 cm`C. `+250cm`D. `+250//9 cm` |
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Answer» Correct Answer - A Focal length `=-(`far point`)` |
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| 560. |
A person who can see things most clearly at a distance of `10 cm`. Requires spectacles to enable to him to see clearly things at a distance of `30 cm.` What should be the focal length of the spectacles ?A. `15 cm` (Concave)B. `15 cm` (Concave)C. `10cm`D. `0` |
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Answer» Correct Answer - A For lens `u=` wants to see `=-30cm` and `v=` can see`=-10cm` `:. (1)/(f)=(1)/(v)-(1)/(u)=(1)/(-10)-(1)/((-30))` `impliesf=-15cm` |
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| 561. |
A person who can see things most clearly at a distance of `10 cm`. Requires spectacles to enable to him to see clearly things at a distance of `30 cm.` What should be the focal length of the spectacles ?A. `15 cm`(Concave)B. `15 cm` (Convex )C. `10 cm`D. `0` |
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Answer» Correct Answer - A For lens `u=`wants to see `=-30cm` and `v=` can see`=-10cm` `:. (1)/(f)=(1)/(v)-(1)/(u)=(1)/(-10)-(1)/((-30))impliesf=-15cm` |
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| 562. |
A man can see clearly up to 3 metres. Prescribes a lens for his spectacles so that he can see clearly up to `12 ` metresA. `-3//4D`B. `3D`C. `-1//4D`D. `-4D` |
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Answer» Correct Answer - C For lens `u=` wants to see`=-12cm` `v=`can see`=-3m` `:. P= (1)/(f)=(1)/(v)-(1)/(u)impliesP=(1)/(-3)-(1)/((-12))=-(1)/(4)D` |
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| 563. |
The proper exposure time for a photographic print is `20 s` at a distance of `0.6 m` from a `40` candles power lamb.How long will you expose the same print at a distance of `1.2 m` from a`20` candles power lamb? |
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Answer» In case of camera, for proper exposure `I_(1)D_(1)^(2)t_(1) = I_(2)D_(2)^(2)t_(2)` As here `D` is constant and `I = (L//r^(2)), (L_(1))/(r_(1)^(2)) xx t_(1)= (L_(2))/(r_(2)^(2))xx t_(2)` So `(40)/((0.6)^(2))xx20 = (20)/((1.2)^(2))t implies t=160"s"` |
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| 564. |
A person uses a lens of power `+3D` to normalise vision. Near point of hypermetropic eye isA. `1m`B. `1.66m`C. `2m`D. `0.66m` |
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Answer» Correct Answer - A Focal length of the lens`f=(100)/(3)` By lens formula `(1)/(f)=(1)/(v)-(1)/(u)` `implies(1)/(+100//3)=(1)/(v)-(1)/(-25)impliesv=-100cm=-1m` |
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| 565. |
A satisfactory photographic print is obtained when the exposure time is `10 sec` at a distance of `2m` from a `60 cd` lamp. The time of exposure required for the same quality print at a distance of `4m` from a `120 cd` lamp isA. `5 sec`B. `10 sec`C. `15 sec`D. `20sec` |
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Answer» Correct Answer - D `I_(1)D_(1)^(2)t_(1)=I_(2)D_(2)^(2)t_(2)` Here `D` is constant and `I=(L)/(r^(2))` So `(L_(1))/(r_(1)^(2))xxt_(1)=(L_(2))/(r_(2)^(2))xxt_(2)` `implies(60)/((2)^(2))xx10=(120)/((4)^(2))xxt=20sec` |
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| 566. |
A point source of light S is placed in front of a perfectly reflecting mirror as shown in Figure. `Sigma` is a screen. The intensity at the center of screen is found to be I. If the mirror is removed, then the intensity at the center of screen would be A. `I`B. `10I//9`C. `9I//10`D. `2I` |
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Answer» Correct Answer - c. Let poewr of light source be P, then intensity at any point on the screen it due to light rays directly received from source and that due to lilght rays after reflection from the mirror. `I=(P)/(4pia^(2))+(P)/(4pixx(3a)^(2))` When mirror is taken away, `I_(1)=(P)/(4pia^(2))=(9I)/(10)` |
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| 567. |
The respective speeds of five molecules are 2,1.5,1.6,1.6 and 1.2 km/sec. The most probable speed in km/sec will beA. `2`B. `1.58`C. `1.6`D. `1.31` |
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Answer» Correct Answer - C Since maximum number of molecules travel with speed `1.6 km//s` so `v_("mp") = 1.6 km//s` |
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| 568. |
What is the least radius through which an optical fiber of core diameter 0.05 mm may be bent (as shown in figure) without serious loss of light? The refractive index of the core is 1.6 and that of cladding is 1.5. A. 0.28 mmB. 0.58 mmC. 0.78mmD. 1 mm |
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Answer» Correct Answer - C `theta_("incident") gt theta_("optical")` `r gt .75` mm |
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| 569. |
The cross section of a prism is a regular hexagon. A narrow beam of light strikes a face of the prism just below the midpoint (M) of the edge AB. The beam is parallel to the top and bottom faces of the prism. Final the minimum value of refractive index of the material of the prism for which the emergent beam will be parallel to the incident beam. |
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Answer» Correct Answer - `mu_(min) = (sqrt(13))/(2)` |
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| 570. |
STATEMENT-`1` A virtual image can be photographed. `STATEMENT 2` Only a real image can be formed on a screenA. Statement is True, Statement II is True, Statement II is correct explanation for Statement IB. Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True. |
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Answer» Correct Answer - b. The ray of light are diverging out from a virtual image. These can be easily converged onto the film of a concave lens by convergent action of its lens. |
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| 571. |
A triangular plate has two cavities , one square and one rectangular as shown in figure . The plate is heated. A. a increase , b decreaseB. a and b both increaseC. a and b increase , x and l decreaseD. a , b , x and l all increase |
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Answer» Correct Answer - B::D All dimensions increase on heating. |
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| 572. |
Letter F is kept in front of a right triangular psim. The light rays enter perpendicular to the large rectangular face, is reflected twice by small rectangular faces and exits perpendicularly to the large rectangular face (see Fig.). Draw the image of the letter seen by the eye. |
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Answer» Correct Answer - E |
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| 573. |
Statement I: A beam of light rays has been reflected from a rough surface. Statement II: Amplitude of incident and reflected rays would be different.A. Statement is True, Statement II is True, Statement II is correct explanation for Statement IB. Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True. |
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Answer» Correct Answer - b. Reflection of light rays takes place on rough as well as smooth dsurface. Some light energy would be absorbed by rough surface, so amplitude of reflected ray is less than that of incident ray. |
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| 574. |
Find the region on Y-axis in which reflected rays are present. Object ia at `A(2,0)` and MN is a plane mirror, as shown in Figure. |
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Answer» Correct Answer - Reflected rays exist on Y-axis between (0,6) and (0,9) |
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| 575. |
See the following figure. Which of the object(s) shown in `O_(2)` figure will not form its image in the mirror. |
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Answer» Correct Answer - `O_(3)` |
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| 576. |
Two plane mirrors are inclined at an angle of `75^(@)` to each other. Find the total number of images formed when an object is placed as shown in figure |
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Answer» Correct Answer - 4 |
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| 577. |
Three plane mirrors are kept as shown in the Figure A point object (O) is kept at the centroid of the triangle seen in the Figure. How many images will be formed? |
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Answer» Correct Answer - 12 |
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| 578. |
The distance between the eye and the feet of a boy is 1.5 m. He is standing on a flat ground and a vertical plane mirror M is placed at a distance of 1.2 m from the boy, with its lower edge at a height of 0.3 m from the ground. Now the mirror is tilted about is lower edge as shown in the Figure. Find maximum value of angle `theta` for which the image of feet remains visible to the boy. [Take `sin 15^(@) = (1)/(4)]` |
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Answer» Correct Answer - `15^(@)` |
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| 579. |
Two plane mirrors are joined together as shown. Two point objects A and B are placed symmetrically such that `OA = OB = d. [AOB` is a straight line] (a) If the images of A and B coincide find `theta` (call it `theta_(0))`. (b) Keeping the position of objects unchanged the angle between the two mirrors is increased to `theta = (4)/(3) theta_(0)`. Now find the distance between the images of A and B. |
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Answer» Correct Answer - (a) `theta_(0) = 90^(@)` (b) d |
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| 580. |
Two plane mirrors M1 and M2 of length d each are placed at right angle as shown. A point object O is placed symmetrically with respect to the mirrors at co-ordinates `((d)/(2),(d)/(2))` (a) How many images of O will be seen? (b) Show that all the images lie on a circle. (c) Length `l (= OA_(1) = OA_(2))` of both the mirrors is cut and removed. Find least value of l such that only two images of the object are formed. |
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Answer» Correct Answer - (a) 3 (c) `l = (d)/(3)` |
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| 581. |
OP is the principal axis of a concave mirror M1. Just below the axis a plane mirror M2 is placed at a distance d from the concave mirror. Two small pins A and B are placed on the principal axis as shown. By moving M2 and changing d, the virtual image of A formed in mirror M1 and the virtual image of B formed in mirror M2 were made to coincide. (a) Calculate the focal length of the concave mirror if it was found that the images coincide when separation between the mirrors was `d_(0)`. (b) Can the two virtual images be observed by the eye simultaneously? |
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Answer» Correct Answer - (a) `|f| = ((x_(2) -d_(0))x_(1))/(x_(2) - (d_(0)+x_(1)))` (b) No. |
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| 582. |
Two plane mirrors M1 and M2 are inclined at `30°` to the vertical. A point object (O) is placed symmetrically between them at a distance of 4 cm from each mirror. Find the distance of the object from the second image formed in mirror M1. |
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Answer» Correct Answer - `8 sqrt(3) cm` |
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| 583. |
O is a small object placed at a distance D from the eye E of an observer. A concave lens of focal length `f (/_D)` is placed near to the eye and image of the object is viewed. Now the lens is moved towards the object O, away from the eye. (a) Show that the angle subtended by the image at the eye first decreases and then increases as the lens is moved away from the eye. (b) Find the distance of the lens from the object when the apparent size of image is smallest. |
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Answer» Correct Answer - (b) `(D)/(2)` |
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| 584. |
The numerical value of the length of Galilean telescope for normal vision is (assuming `f_0` and `f_e` as positive length)A. `f_(0)xxf_(e)`B. `(f_(0))/(f_(e))`C. `f_(0)+f_(e)`D. `f_(0)-f_(e)` |
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Answer» Correct Answer - D `f_(0)-f_(theta)` |
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| 585. |
A concave mirror has a focal length of 20 cm. Find the position or positions of an object for which the image-size is double of the object-size. |
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Answer» Correct Answer - A::C For the concave mirror, `f=-20cm` `M=v/u=2` `rarr v=-2u` Ist case (virtual image) `1/v+1/u=1/f` `rarr 1/(2u)1/u=1/f` `rarr 1/(2u)=1/f` `rarr u=f/2=10cm` `:.` The positions are 10 cm or 30 cm from the concave mirror. |
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| 586. |
A man standing on sea-shore sees an elongated image (shown by dashed line) of a floating object AB. In fact he finds the image to be oscillating due to air turbu- lence. Figure (ii) gives three plots (a, b and c) of height from the water surface vs air temperature. Which one best illustrates the air-temperature condition that can create this image? [Many people have seen sea monsters due to this phenomena!] |
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Answer» Correct Answer - C |
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| 587. |
A man is Standing on the peak of a mountain and finds that evening sun rays are nearly horizontal. At a horizontal distance of 6 km from him, its raining and he sees a beautiful rainbow. The sun rays entering water drop get refracted, reflected and refracted to form a rainbow. The red light is emitted from a drop upto a maximum angle of `42°` with respect to the incident sunlight. In front of the man there is a flat valley at a depth of 0.5 km from the mountain peak. What fraction of the complete circular arc of the rainbow is visible to the man? |
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Answer» Correct Answer - 0.53 |
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| 588. |
A man of height 1.8 m is standing in front of a wall. The sun is exactly behind him. His shadow has a length 1.5 m on the ground and 0.75 m on the wall. Find the length of his shadow on the ground if the wall is removed. |
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Answer» Correct Answer - `2.57 m ` |
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| 589. |
Find the minimum size of mirror required to see the full image of a wall behind a man standing at the centre of room, where H is the height of wall. |
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Answer» Correct Answer - `H//3` |
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| 590. |
The focal lengths of the objective and the eye piece of a compound microscope are 1cm and 5cm respectively. An object placed at a distance of `1:1` cm from the objective has its final image formed at (i) infinity (ii) least distance of distinct vision. Find the magnifying power and the distance between the lenses. Least distance of distinct vision is 25 cm. |
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Answer» Correct Answer - (i) 16 cm, 50 cm; (ii) 15.17 cm, - 60 |
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| 591. |
The focal lengths of the objective and eye`-` lens of a microscope are `1cm` and `5 cm` respectively. If the magnifying power for the relaxed eye is `45`, then the length of the tube isA. `6 cm`B. `9 cm`C. `12 cm`D. `15 cm` |
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Answer» Correct Answer - D Magnifying power for relaxed eye is `m=(v_(o))/(u_(o))(D)/(f_(e))` Here, `M=-45,` `f_(o)=1 cm` `f_(e)=5 cm,` `D=25 cm` `:. -45 =-(v_(o))/(u_(o)). (25)/(5),` or `(v_(o))/(u_(o))=9,` For objective image is real. Therefore, `u_(o)=v_(o)` `u_(o)=-(v_(o))/(9),f_(o)=+1cm` Substituting in `(1)/(v_(o))-(1)/(u_(o))=(1)/(f_(o)),` we get `(1)/(v_(o))+(9)/(v_(o))=1` or`v_(o)=10cm` `:.` Length of tube `L=v_(o)+f_(e)=10+5=15cm` |
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| 592. |
STATEMENT-1: A lens has two principal focal lengths which may be different in magnitude. STATEMENT-2: The distance of both principal focus from optical centre of lens depend on the two radii of curvature of the lens. Distance of both principal focus from optical centre a lens are same only if radii of curvature of both sides of lens are sameA. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - C A lense will both focal to be different, if medium on both sides of lense have different refractive indices. Hence statement 1 is true statement 2 is false. |
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| 593. |
Rays of light from Sunn falls on a biconvex lens of focal length f an the circular image of Sun of radius r is formed on the focal plane of the lens. Then,A. area of image is `pir^(2)` and area is directly proportinal of fB. area of image is `pir^(2)` and area is directly proportional to `t^(2)`C. intensity of image increases if f is increasesD. If lower half of the lens is covered with black paper, area will become half |
| Answer» Correct Answer - b. | |
| 594. |
A parallel beam of white light falls on a combination of a concave and a convex lens, both of the same material. Their foacl lengths are 15cm and 30cm, respectively for the mean wavelength in white light. On the other side of the lens system, one sees colored patterns with violet color at the outer edge. |
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Answer» True. `(1)/(F)=(1)/(f_(1))+(1)/(f_(2))rArr (1)/(F)=(1)/(-15)+(1)/(30)` `=(-2+1)/(30)rArr F=-30cm. ` This combination behaves as a concave lens of focal length 30 cm. Since `F_(v) lt F` , therefore one sees colored pattern with violet color at the outer edge. the statement is true. |
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| 595. |
When a ray of light enters a glass slab from air.A. its wavelength decreasesB. its wavelength increasesC. its frequency increasesD. neither its wavelength nor its frequency changes |
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Answer» Correct Answer - a. `lambda=(v)/(f)` In moving from air to glass, f remains unchanges while v decreases. Hence, `lambda` should decrease. |
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| 596. |
A convex lens of focal length 40cm is in contact with a concave lens of focal length 25cm. The power of the combination isA. `-1.5` dioptresB. `-6.5` dioptresC. `+6.5` dioptresD. `+6.67` dioptres |
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Answer» Correct Answer - A `(1)/(f)=(1)/(f_(1))+(1)/(f_(2))=(1)/(0.4)+(1)/(-0.25)=(1)/(0.4)-(1)/(0.25)` `(0.25-0.4)/(0.4xx0.25)=(-0.15)/(0.4xx0.25)=(-0.15)/(0.4xx0.25)=-1.5` `rArrP=(1)/(f)=-1.5D` |
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| 597. |
A thin prism having refracting angle `10^(@)` is made of glass of refracting index `1.42`. This prism is combined with another thin prism of glass of refractive index `1.7`. This combination produces dispersion without deviation. The refracting angle of second prism should be `:`A. `6^(@)`B. `8^(@)`C. `10^(@)`D. `4^(@)` |
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Answer» Correct Answer - A For dispersion without deviation `delta_(1)=delta_(2)` `A_(1)(mu_(1)-1)=A_(2)(mu_(2)-1)` `10(1.42-1)=A_(2)(1.7-1)` `A_(2)=6^(@)` |
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| 598. |
A Carnot engine takes `3xx10^6` cal of heat from a reservoir at `627^@C` and gives it to a sink at `27^@C`. The work done by the engine is:A. `4.2 xx 10^(6)J`B. `8.4 xx 10^(6)J`C. `16.8 xx 10^(6)J`D. zero |
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Answer» Correct Answer - B `T_("source") = 627^(@)C = 627 + 273 = 900K` `T_("sink") = 27^(@)C = 27 + 273 = 300K` Efficiency `(eta)= 1 - (T_("sink"))/(T_("source")) = 1-(300)/(900) = 1- (1)/(3)` `(eta) = (2)/(3) = ("Output")/("Input") = ("Work")/(Heat Input")` `implies (2)/(3) xx "Heat Input" = "Work"` `implies (2)/(3) xx 3xx10^(6) xx 4.2 = "Work"` `implies "Work" = 8.4 xx 10^(6)J` |
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| 599. |
The specific heat of argon at constant volume is `0.075 kcal kg^(-1) K^(-1)`. Calculate its atomic weight. `Take R = 2 cal mol^(-1) K^(-1)`. |
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Answer» As argon is monoatomic ,its molar specific heat at constant volume will be `C_(V) = (3)/(2)R = (3)/(2) xx 2 = 3 "cal" // "mol"K, C_(V) = M_(w)c_(v)` `" "` and `c_(v)= 0.075"cal"//gK` So `3 = M_(w) xx 0.075 implies M_(w) = (3)/(0.075) = 40"gram"//"mole"` Effecieny of a cycle `(eta)`: `eta = ("total Mechanical work done by the gas in the whole process")/("Heat absorbed by the gas (only +ve)") = ("area under the cycle in " P-V " curve")/("Heat injected into the system")` |
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| 600. |
A carnot engine working between `400K` and `800K` has a work output of `1200 J` per cycle . What is the amount of heat energy supplied to the engine from source per cycle? |
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Answer» `W = 1200 J, T_(1) = 8000k, T_(2) = 400 K` `therefore " " eta = 1 - (T_(2))/(T_(1)) = (W)/(Q_(1)) implies 1-(400)/(800) = (1200)/(Q_(1))` " " 0.5 =(1200)/(Q_(1))` Heat energy supplied by source `Q_(1) = (1200)/(0.5) = 2400` joule per cycle |
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