InterviewSolution
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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.
| 1. |
Can you ever have a situation in which a light ray goes undeviated through a prism? |
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Answer» If the light ray falls perpendicularly on the surface and the angle of the prism is zero, i.e. i=i'=A=0 →ẟ = i+i'-A = 0 But in this case the sides of the prism are parallel and we may not call it a prism practically because it will look like a cuboid. Another case is when the prism is submerged in a dense liquid which refractive index is the same as the prism. Since the speed of light will not change in both medium hence the light will travel undeviated through the prism. |
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| 2. |
Two concave lenses L1 and L2 are kept in contact with each other. If the space between the two lenses is filled with a material of refractive index μ=1 the magnitude of the focal length of the combination(a) becomes undefined(b) remains unchanged(c) increases(d) decreases |
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Answer» (c) increases EXPLANATION: The focal length F of the combination is given as 1/F = 1/f₁ + 1/f₂ It is like resistors connected in parallel. The numerical value of equivalent resistance is less than the smallest resistor. Similarly, the numerical value of the equivalent focal length will decrease. Since the 1/F is the power of the lens and in the case of a diverging lens, it is negative as in this case. Hence F is also negative. If the numerical value of F decreases that means its negative will increase. |
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| 3. |
Consider three converging lenses L1, L2 and L3, having identical geometrical construction. The index of refraction of L1 and L2 are μ1, and μ2, respectively. The upper half of the lens L3 has a refractive index and the lower half has μ2 (figure 18-Q4). A point object O is imaged at O1, by the lens L2, and at O2 by the lens L2 placed in same position. If L3 i s placed at the same place,(a) there will be an image at O1,(b) there will be an image at O2.(c) the only image will form somewhere between O1 and O2,(d) the only image will form away from O2. |
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Answer» (a) there will be an image at O1 (b) there will be an image at O2. |
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| 4. |
A thin lens is made with a material having a refractive index µ = 1.5. Both the sides are convex. It is dipped in water (µ = 1.33). It will behave like (a) a convergent lens (b) a divergent lens (c) a rectangular slab (d) a prism |
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Answer» (a) a convergent lens EXPLANATION: The double convex lens behaves as a convergent lens in the air because the µ of the lens is greater than air. When it is dipped in water its µ is still greater than water hence it will act as a convergent lens. |
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| 5. |
Is the formula "Real depth/Apparent depth =µ" valid if viewed from a position quite away from the normal? |
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Answer» No. Because in its derivation the formula Sin i/sin r = µ has been used and keeping the angles i and r very small, the approximation sin i ≈ tan i, and sin r ≈tan r have been taken. When viewed from a position quite away from the normal these approximations do not hold good. Hence the formula "Real depth/ Apparent depth=µ" is not valid. |
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| 6. |
Mark the correct options.(a) If the incident rays are converging, we have a real object.(b) If the final rays are converging, we have a real image.(c) The image of a virtual object is called a virtual image.(d) If the image is virtual, the corresponding object is called a virtual object. |
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Answer» (b) If the final rays are converging, we have a real image. Explanation: This is because a real image is formed by converging reflected/refracted rays from a mirror/lens |
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| 7. |
If the light moving in a straight line bends by a small but fixed angle, it may be a case of(a) reflection(b) refraction(c) diffraction(d) dispersion. |
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Answer» (a) reflection (b) refraction |
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| 8. |
A convex lens is made of a material having refractive index 1.2. Both the surfaces of the lens are convex. If it is dipped into water (μ = 1'33), it will behave like(a) a convergent lens(b) a divergent lens(c) a rectangular slab(d) a prism. |
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Answer» (b) a divergent lens EXPLANATION: The µ of the convex lens is greater than air so it will act as a convergent lens in the air. When it is dipped in water its µ is less than water. Now the rays entering and exiting the lens will bend toward or away from the normal in an opposite fashion than when the lens was placed in air. So the convergent lens in the air will now behave as a divergent lens in the water. |
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| 9. |
The image formed by a concave mirror(a) is always real(b) is always virtual(c) is certainly real if the object is virtual(d) is certainly virtual if the object is real. |
| Answer» (c) is certainly real if the object is virtual | |
| 10. |
A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at(a) infinity(b) pole(c) focus(d) 15 cm behind the mirror. |
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Answer» (d) 15 cm behind the mirror. EXPLANATION: The option can be selected by suitable reasoning. In a convex mirror, the object at infinity has its image at the focus behind the mirror. So option (c) is not correct because the object here is not at infinity. The image in a convex mirror never forms at infinity. The option (a) is not correct. The image will be near the pole if the object is near the pole, which is not the case here. The option (b) is not correct. The only option left is (d) and the image position is between focal length and the pole for the object between infinity and the pole. Hence the option (d) is correct. |
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| 11. |
The image of an extended object, placed perpendicular to the principal axis of a mirror, will be erect if(a) the object and the image are both real(b) the object and the image are both virtual(c) the object is real but the image is virtual(d) the object is virtual but the image is real. |
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Answer» (c) the object is real but the image is virtual (d) the object is virtual but the image is real. |
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| 12. |
A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light(a) remains, constant(b) continuously increases(c) continuously decreases(d) first increases then decreases. |
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Answer» (d) first increases then decreases. EXPLANATION: The parallel beam of light converges at the focus on the other side and then diverges. Hence till the person moves to the focus the intensity will increase. Beyond the focus, it will decrease. Hence the option (d). |
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| 13. |
The equation of refraction at a spherical surface is µ₂/v - µ₁/u =(µ₂-µ₁)/R Taking R = ∞, show that the equation leads to the equation Real depth/Apparent depth = µ₁/µ₂ for refraction at a plane surface. |
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Answer» The equation of refraction at a spherical surface is µ₂/v - µ₁/u =(µ₂-µ₁)/R Given that R = ∞, (Implies that the surface is plane) Hence, µ₂/v - µ₁/u = 0 →µ₂/v = µ₁/u →µ₁/µ₂ = u/v = h₁/h₂ = Real Depth/Apparent depth |
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| 14. |
Figure (18-Q3) shows three transparent media of refractive indices μ1, μ2 and μ3, A point object o is placed in the medium μ2. If the entire medium on the right of the spherical surface has refractive index μ1, the image forms at o'. If this entire medium has refractive index μ3, the image forms at o'. In the situation shown,(a) the image forms between o' and o"(b) the image forms to the left of o'(c) the image forms to the right of o"(d) two images form, one at o' and the other at o". |
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Answer» (d) two images form, one at o' and the other at o". EXPLANATION: The rays incident on the surface above the principal axis will be refracted in the medium of R.I. µ₃ and form the image at O". But the rays falling below the axis will be refracted in the medium having R.I. µ₁ and the image will be formed at O'. Hence two images will be formed. |
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| 15. |
A point source of light is placed at a distance, of 2 f from a converging lens of focal length 1 The intensity on the other side of the lens is maximum at a distance(a) f(b) between f and 2 f (c) 2 f (d) more than 2 f. |
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Answer» The correct answer is (c) 2 f EXPLANATION: The intensity of the light on the other side is maximum where the image of the point source is formed. Here u = -2f, f = f, hence from the lens formula 1/v -1/(-2f) = 1/f →1/v = 1/f -1/2f = 1/2f →v = 2f Hence the image will be formed at 2f distance from the lens on the other side. |
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| 16. |
Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct ?(a)1/v-1/u=t/uf(b) t/v2-1/u=1/f(c) 1/(v-t)- 1/(u+t) =1/f(d) 1/v-1/v+ t/uv= t/f. |
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Answer» (c) 1/(v-t)- 1/(u+t) =1/f EXPLANATION: The thickness of the lens will affect the object distance or image distance. Either the thickness t will be added or subtracted to u or v. It will not be in a proportion as shown in the options (a), (b) and (d). Hence the option (c) is correct. |
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| 17. |
A double convex lens has two surfaces of equal radii R and refractive index μ= 1.5. We have,(a) f= R/2(b) f = R(c) f = - R(d) f =2R. |
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Answer» The correct answer is (b) f = R EXPLANATION: From lens makers formula 1/f = (µ-1){1/R - 1/R'} here µ = 1.5, R' = -R →1/f = 0.5*2/R =1/R →f = R Hence the option (b) is true. |
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| 18. |
Find the maximum angle of refraction when a light ray is refracted from glass (μ= 1.50) to air. |
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Answer» From the definition of critical angle, if refracted angle is more than 90°, then reflection occurs, which is known as total internal reflection. So, maximum angle of refraction is 90°. |
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| 19. |
Light falls from glass (μ= 1.5) to air. Find the angle of incidence for which the angle of deviation is 90°. |
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Answer» Since, μ= 1.5, Critial angle = sin–1(1/μ) = sin–1 (1/1.5) = 41.8° |
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| 20. |
Can mirrors give rise to chromatic aberration? |
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Answer» In mirrors laws of reflection are followed. The angle of incidence and the angle of reflection are the same for each wavelength of the light. Hence the focal length of a mirror is the same for each wavelength of the light. Thus mirrors can not give rise to chromatic aberration. |
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| 21. |
A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens(a) must be less than 10 cm(b) must be greater than 20 cm(c) must not be greater than 20 cm(d) must not be less than 10 cm. |
| Answer» (a) must be less than 10 cm | |
| 22. |
If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object? |
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Answer» If the object is far away from a convex mirror the virtual image is near the focus. As the object moves towards the mirror the image moves towards the back of the mirror. So in an equal interval of time, the object moves a larger distance than the image. So the image moves slower than the object. |
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| 23. |
In motor vehicles, a convex mirror is attached near the driver's seat to give him the view of the traffic behind. What is the special function of this convex mirror which a plane mirror cannot do? |
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Answer» A convex mirror forms a virtual, erect and diminished image behind the mirror if the object is in front of it. So it covers a vast area behind the driver. The plane mirror also forms virtual and erect image behind the mirror but it is of the same size. So the plane mirror does not cover a wide field behind the driver. Hence the convex mirror is used. |
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| 24. |
Why does a diamond shine more than a glass piece cut to the same shape? |
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Answer» There are two reasons. First, due to the higher refractive index, the critical angle for diamond is less and the light goes a larger number of total internal reflections inside the diamond than the glass. Second, there is more variation of µ between violet and red light for the diamond than the glass. Hence the dispersion is wider in diamond than the glass. These two reasons combined make the diamond more shiny and sparkling. |
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| 25. |
A narrow beam of light passes through a slab obliquely and is then received by an eye (figure 18-Q1). The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation. |
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Answer» When the index of refraction, µ changes the refracted rays in the slab change its direction. Thus they emerge out of the slab at different points. This results in the shift of the image of the source and the eye find the source disappearing/shifting a bit with time and the source appears twinkling like a star. |
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