Physics - Mr. Sreekumaran Nair
1.Ans:Wavelength
2. Ans:Photo electric effect.
3. Ans:4NL.
4. Ans: It will turn gradually blue.
5. Ans:Zero.
6. Ans:90 ͦ.
7. Ans:μo= μe
8. Ans:90 ͦ
9. Ans:Violet.
10. Ans: Shrinks.
11. Ans.(i) The upper part of the mirror is convex.
(ii) The middle part of the mirror is concave.
(iii) The lower part of the mirror is plane.
12. Ans. The ray of light bends away from the normal.
13. Ans.
As m = sini/sinr = c/v or v = sinr/sini *c
For a given angle of incidence, v ∝ sinr, vA ∝sin15° , vB ∝ sin25° , vC ∝ sin35°
But sin15° < sin25°< sin35°.
∴ vA < vB < vC .
i.e. the velocity of light is minimum in medium A.
14 Ans.For i = 90°, lateral shift is maximum and is equal to the thickness of the slab.
d = t sin( i – r )/ cos r
dmax = t sin(90°- r)/cos r = t cos r/cos r = t.
15. Ans.The apparent shift caused by a slab of thickness ‘t’ is given by
d = t( 1 -1/μ )
As the refractive index of the galass is maximum for red light, so red coloured letters are
more raised up.
16. Ans. No Apparent depth is maximum for that part of the bottom of the tank which is observed normally. Apparent depth decreases with increasing obliquity. Due to this unequal refraction, the flat bottom of the tank appears concave.
17. Ans. For glass-air interface, sin ic = 1/aμg
The critical angle i’c for glass water interface is given by
Sin i’c = 1/ wμg
Now wμg < aμg
Sin i’c > Sin ic or i’c > ic
18. Ans. Light entering water is totally reflected from the air bubble. For the observer, this light appears to come from the buble. So it shines.
19. Ans. As the critical angle for diamond-oil interface is greater than that for the diamond – air interface, so the shining of diamond reduces when it is dipped in a transparent oil.
20. Ans. It behaves like a biconvex lens.
21. Ans. Air bubble has spherical surface and is surrounded by medium ( water) of higher refractive index. When light passes from water to air it gets diverged. So air bubble behaves as a concave lens.
22. Ans.When the refractive index of the liquid is same as the lens material, no light will be reflected by the lens and hence it will not be visible. µ
23. Ans. No, the image will be formed at the same position.From lens maker’s formula, 1/f = (μ -1) [ 1/R1 –1/R2 ] , it is clear that when we interchangeR1 and R2, the magnitude of‘f’ remains the same
24. Ans. focal length‘f’ of a convex lens is related to its refractive index as
f ∝ 1/(μ -1)
As wμg < aμg , so focal length of a convex lens will increase when it is immersed in water.
5. Ans.Focal length, f ∝ 1/(μ -1) As μR < μV , so the focal length of a convex lens will increase when red light is used.
26. Ans:For the original lens: R1 = +R and R2 = -R, so we can write
1/f = ( μ -1) [ 1/R +1/R ] = 2( μ -1)/R.
When one surface is made plane by grounding, we have R1 = +R and R2 = -∞
Therefore, 1/f ’ = ( μ -1) [ 1/R +1/∞ ] =( μ -1)/R
∴ f ‘ / f = 2 or f ‘ = 2f
Thus the focal length becomes double and power becomes one –half.
27. Ans. When the prism is held in water,
wμg = Sin (A + dm/2) /SinA/2
As wμg < aμg , so the angle of minimum deviation decreases in water.
28. Ans. Total internal reflection.
29. Ans. The sunlight will not be scattered in the absence of atmosphere. So the sky will appear dark.
30. Ans. Clouds have large particles like dust and water droplets which scatter light of all colours almost equally, hence clouds generally appear white.
31Ans. When the sun or the moon is seen through a thin veil of high clouds, holes are seen. These are formed due to reflection of light by the icy crystals present in the atmosphere.
32. Ans. Ultra-violet light has wavelength shorter than that of violet light. Bees have some retinal cones that are sensitive to ultra violet light, so they can see objects in ultra-violet light. Human eyes do not posses retinal cones sensitive to ultra-violet light, so human beings cannot see objects in ultra-violet light. In other words, human beings are ultra-violet blind.
33. Ans. In a chicken’s eye, the retina has a large number of cones but only few rods. The rods are sensitive to bright light only. That is why a chicken is not able to see in dim light. As it needs bright light to see, so it wakes up early in the morning with the sunrise and goes to sleep by sunset.
34. Ans. Magnifying power of a simple microscope ,
m = 1 + D/f
as fV < fR so the magnifying power is greater when the object is seen in violet light.
35. Ans. This is done so that the objective lens forms image within the focal length of the eyepiece.
36. Ans. (i) We should take f0 =1 cm and fe = 3cm for a microscope.
(ii) We should take f0 = 100 cm and fe = 1 cm for a telescope.
37. Ans. Yes, because the light gathering power of objective will increase and even faint objects will become visible.
38. Ans. For relaxed eye,
L = f0 + fe (normal adjustment)
For least distance of distinct vision,
L’ = f0 + ue , ue < fe
Therefore, L’ < L. so that distance between the two lenses should be decreased.
39. Ans. Here u = - (f+a), v = -(f+b), f = -f
As 1/f = 1/u +1/v
F = uv/ u + v
Or -f = [-(f+a)] x [-(f+b)] / -(f+a) – f(a+b)
= f2 + af +bf +ab / -(2f+a+b)
or 2 f2 + af +bf = f2 + af +bf +ab
or f2 = ab
40.Ans. (i) Angle of refraction (θ/2) in medium 2 is less than the angle of incidence (θ) in
medoum 1 i.e. the ray bends towards the normal in medium 2. so medium 2 is optically
denser than medium 1.
(ii) From Snell’s law,
m = sin i/sin r = sin θ/ sin θ/2 = 2sin θ/2 cos θ/2 /sin θ/2 = 2 cos θ/2
Also μ = c1 / c2
hence 2 cos θ/2 = c1 /c2 or θ = 2cos-1(c1 / 2c2) .
41. Ans. The point of convergence shifts away from the glass, as shown in the ray diagram given below. The screen has to be moved towards right to receive the point of convergence again.
42. Ans. Real depth = y cm
Apparent depth = y- x cm
Refractive index of oil,
μ = real depth/ apparent depth = y / y-x
43. Ans. Using Snell’s Law for refraction from glass to air,
Sin i/sin r = gμa = v / c
Where c is the speed of light in air and v is the speed of light in glass, In the consition of
critical incidence, we have
i = ic and r = 900
Sin ic/ sin 900 = v / c or Sin ic
= v / c
Or ic = sin -1 ( v / c)
44. Ans. Twinkling of stars. The light from stars undergoes refraction continuously before it
reaches earth. So the apparent position of the stars is slightly different than its actual position.
Due to variation in atmosphere conditions, like change in temperature, density etc., and this
apparent position keeps on changing. The amount of light entering our eyes from a particular
star increases and decreases randomly with time. Sometimes, the star appears brighter and
other times, it appears fainter. This gives rise to the twinkling effect of stars.
The planets do not show twinkling effect. As the planets are much closer to the earth, the
greater and the fluctuations caused in the amount of light due to atmospheric refraction are
negligible as compared to the amount of light received from them.
45. Ans. Light from the stars near the horizon reaches the earth obliquely through the atmosphere. Its path changes due to refraction. Frequent atmospheric disturbances change the path of light and cause twinkling of stars. Light from the stars overhead reaches the earth normally. It does not suffer refraction. There is no change in its path. Hence there is no Twinkling effect
46. Ans. Magnification produced by any lens,
m = v/u = f / f + u
given m = ± N ±N = f / f + u
or f + u = ± f / N or u = - f ± f / N
hence magnitude of object distances,
|u| = f ± f / N
given P = 1/f = + 2.5 D
f = 1/ 2.5 = 0.4 m = 40 cm
Also N = 4
|u| = 40 ± 40/4 = 40 ± 10 = 50 cm or 30 cm.
(a) for a convex lens, f>>0 and for an object on left, u<0. when the object is placed within the
focus of a convex lens,
0 < |u| < f or 0< 1 / |u| > 1/f
1/v = 1/f+1/u=1/f-1/|u|<0
i.e. v < 0 so a virtual image is formed on left.
Now as u<0 and v<0, so 1/v = 1/f + 1/u
= - 1/ |v| = 1/f – 1/|u| or 1/|u| - 1/|v| = 1/f
As f>0
1/|u| - 1/|v| > 0 or 1/|u| > 1/|v| or |u|<|v|
i.e. |v|>|u| |m| = |v/u| > 1
Hence image is enlarged.
(b) For a concave lens f<0 and for an object on left, u<0
1/v = 1/f +1/u = 1/|f| - 1/|u|
= - [1/|f|+1/|u|] < 0 for all u.
i.e. v<0 for all values of u. hence a virtual image is formed on the left.
Also 1/|v| = 1/|f| + 1/|u| 1/|v| > 1/|u|
Or |v|<|u| |m| = |v/u| < 1
i.e. the image is diminished in size.
48. Ans. A hollow prism contains air which does not cause dispersion. The faes AB and AC of the hollow prism behave like parallel sides of glass plates. The beam is laterally deviated at each of the two refracting faces. However, the rays of different colours emerge parallel to each other. So there is no dispersion.
49. Ans. (i) Moon has no atmosphere. There is no scattering of light. Sunlight reaches moon straight covering shortest distance. Hence sunrise and sunset are abrupt. (ii) Moon has no atmosphere. So there is nothing to scatter sunlight towards the moon. No skylight reaches moon surface. Sky appears black in the day time as it does at night. (iii) No water vapours are present at moon surface. No clouds are formed. There are no rains on the moon. So rainbow is never observed.
50. Ans. The total apparent shift in the position of the image due to all the three media is
given by
d = t1[1-1/(μ1)+ t2[1-1/(μ2)+ t3[1-1/(μ3)
Given t1 = 4.0 cm,t2=6.0 cm , t3 =8.0 cm
μ1 = 1.5 , μ2=1.4 , μ3 =1.3 cm
d= 4.0(1-1/1.5)+6.0(1-1/1.4)+8.0(1-1/1.3)
= 1.33 + 1.71 + 1.85 = 4.89 cm
51. Ans. Clearly , the fish can see the outside view of the cone with semi vertical angle,
But m = 1 / sin ic
Or 1/3 = 1 / sin ic
Or sin ic = ¾ = 0.75
q/2 = ic = sin-1 (0.75) = 48.60
52. Ans. (i) As the lens forms a real iamge, it must be a convex lens.
(ii) From the graph, when u=20 cm , we have v = 20 cm.
For the convex lens forming a real iamge, u is negative and v and f are positive.
U = -20 cm v = +20cm
Using this lens formula,
1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm
53. Ans. A=600 , δm=300
i = e = ¾ A = 450 ,
as A + δ = i + e
60 + δ = 45 + 45
or δ = 300
Refractive index,
m = sin a+δm /2 /sin A/2 = sin 600+300/2 / sin 600/2
= sin 450/sin300 = 1/√2 / ½ = √2 = 1.414
54. Ans. Two light sources will be coherent if (i) The frequency of the two light sources is same and, (ii) The phase difference between them remains constant.
55. Ans. Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms, when they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.
56. Ans. Fringe width , β = λD/d
i.e. β ∝ 1/d , when d →0, β → ∞
fringe width is very large. Even a single fringe may occupy the entire screen. The
interference pattern cannot be observed.
57. Ans. The given path difference satisfies the condition for the minimum of intensity for yellow light, Hence if yellow light is used, a dark fringe will be formed at the given point. If white light is used, all components of white light except the yellow one would be present at this point.
58. Ans. The positions of bright and dark fringes will change rapidly. Such rapid changes cannot be detected by our eyes. A uniform illumination is seen on the screen i.e. interference pattern disappears.
59. Ans. For diffraction to take place the wave length should be of the order of the size of the obstacle. The radio waves (particularly short radio waves) have wave length of the order of the size of the building and other obstacles coming in their way and hence they easily get diffracted. Since wavelength of the light waves is very small. They are not diffracted by the buildings.
60. Ans. Muslin cloth is mde of very fine threads and as such fine slits are formed. White light passing through these silts gets diffracted giving rise to colored spectrum. The central maximum is white while the secondary maxima are coloured. This is because the positions of secondary maxima (except central maximum) depend on the wavelength of light. In a coarse cloth, the slits formed between the threads are wider and the diffraction is not so pronounced. Hence no such spectrum is seen.
CBSE Physics (Chapter Wise With Hint / Solution) Class XII (By Mr. Sreekumaran Nair)