XII Physics Question Bank-pages-28

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Previous Year CBSE Questions Page 271 of 484 
 
at I1. Since image I1 is real, it serves as a virtual object for the second lens B, producing the final 
image at I. 
For the image formed by the first lens A, we get 
1
𝑣1
−
1
𝑢
=
1
𝑓1
…… . (1) 
For the image formed by the second lens B, we get 
1
𝑣
−
1
𝑣1
=
1
𝑓2
…… (2) 
Adding Eqs. (1) and (2), we get 
1
𝑣
−
1
𝑢
=
1
𝑓1
+
1
𝑓2
……(3) 
If the two lens-system is regarded as equivalent to a single lens of focal length f we get 
1
𝑣
−
1
𝑢
=
1
𝑓
……(4) 
From (3) and (4) 
1
𝑓1
+
1
𝑓2
=
1
𝑓
 
62. Refer question 4 of section A 5 marks questions. 
63. (i) n1 = n2 
 
 
 
(ii) n1 n2 
 
 
 
 
SOLUTION OF 5 MARKS QUESTIONS 
 
64. (a) Ray diagram: 
(b) The two right- angled triangles 𝐴′𝐵′𝐹and MPF 
are similar. 
𝐵′𝐴′
𝑃𝑀
=
𝐵′𝐹
𝐹𝑃
 
𝐵′𝐴′
𝐵𝐴
=
𝐵′𝐹
𝐹𝑃
(∵ 𝑃𝑀 = 𝐵𝐴)………(1) 
∠𝐴𝑃𝐵 = ∠𝐴′𝑃𝐵′ 
The right triangles 𝐴′𝐵′𝑃 and ABP are also similar 
𝐵′𝐴′
𝐵𝐴
=
𝐵′𝑃
𝐵𝑃
……(2) 
Comparing eqs. (1) and (2) 
𝐵′𝐹
𝐹𝑃
=
𝐵′𝑃
𝐵𝑃
 
𝐵′𝑃 − 𝐹𝑃
𝐹𝑃
=
𝐵′𝑃
𝐵𝑃
…… . (3) 
Taking the sign conventions, 𝐵′𝑃 = −𝑣, 𝐹𝑃 = −𝑓, 𝐵𝑃 = −𝑢 
Using these in equation (3) 
−𝑣 + 𝑓
−𝑓
=
−𝑣
−𝑢
 
𝑣 − 𝑓
𝑓
=
𝑣
𝑢
 ⟹ 
𝑣
𝑓
− 1 =
𝑣
𝑢
 ⟹ 
1
𝑓
−
1
𝑣
=
1
𝑢
……(𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 𝑣) 
∴
1
𝑓
=
1
𝑣
+
1
𝑢
 
The above equation is mirror formula. 
Linear magnification: 𝑚 =
ℎ′
ℎ
 
Previous Year CBSE Questions Page 273 of 484 
 
From eqn (2) 
𝐵′𝐴′
𝐵𝐴
=
𝐵′𝑃
𝐵𝑃
 ∴ 
−ℎ′
ℎ
= 
−𝑣
−𝑢
 
𝑚 =
ℎ′
ℎ
= −
𝑣
𝑢
 
(c) The two advantages of reflecting telescope: 
(i)There is no chromatic aberration in case of reflecting telescope as the objective is a mirror. 
(ii) Spherical aberration is reduced in case of reflecting telescope by using mirror objective in the 
form of paraboloidal mirror. 
 
65. The figure shows the geometry of image 
formation of image I of an object O on the principal 
axis of a spherical surface with centre of curvature C, 
and radius of curvature R. 
 
Consider the aperture of surface to be small compared 
to other distances involved, so that the angle 
approximation can be made. 
𝑡𝑎𝑛∠𝑁𝑂𝑀 ≈ ∠𝑁𝑂𝑀 = 
𝑀𝑁
𝑂𝑀
 
𝑡𝑎𝑛∠𝑁𝐶𝑀 ≈ ∠𝑁𝐶𝑀 = 
𝑀𝑁
𝑀𝐶
 
𝑡𝑎𝑛∠𝑁𝐼𝑀 ≈ ∠𝑁𝐼𝑀 = 
𝑀𝑁
𝑀𝐼
 
𝐹𝑜𝑟 ∆𝑁𝑂𝐶, ′𝑖′ 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒. 
∴ 𝑖 = ∠𝑁𝑂𝑀 + ∠𝑁𝐶𝑀 
 
∴ 𝑖 = 
𝑀𝑁
𝑂𝑀
+
𝑀𝑁
𝑀𝐶
…… . (1) 
 
𝐹𝑜𝑟 ∆𝑁𝐶𝐼, ∠𝑁𝐶𝑀 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒. 
∴ 𝑟 = ∠𝑁𝐶𝑀 − ∠𝑁𝐼𝑀 
 
∴ 𝑟 = 
𝑀𝑁
𝑀𝐶
−
𝑀𝑁
𝑀𝐼
…… . (2) 
 
By Snell’s law 
𝑛1 sin 𝑖 = 𝑛2 sin 𝑟 
 
Previous Year CBSE Questions Page 274 of 484 
 
𝑛1𝑖 = 𝑛2𝑟…… (𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑎𝑛𝑔𝑙𝑒𝑠) 
 
𝑛1 (
𝑀𝑁
𝑂𝑀
+
𝑀𝑁
𝑀𝐶
) = 𝑛2 (
𝑀𝑁
𝑀𝐶
−
𝑀𝑁
𝑀𝐼
) …… (𝑓𝑟𝑜𝑚 (1)&(2)) 
 
𝑛1
𝑂𝑀
+
𝑛1
𝑀𝐶
=
𝑛2
𝑀𝐶
−
𝑛2
𝑀𝐼
………(𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 𝑀𝑁) 
 
𝑛1
𝑂𝑀
+
𝑛2
𝑀𝐼
=
𝑛2 − 𝑛1
𝑀𝐶
 
 
Applying the sign convention, 𝑂𝑀 = −𝑢, 𝑀𝐼 = +𝑣, 𝑀𝐶 = +𝑅 
𝑛1
−𝑢
+
𝑛2
𝑣
=
𝑛2 − 𝑛1
𝑅
 
𝑛2
𝑣
−
𝑛1
𝑢
=
𝑛2 − 𝑛1
𝑅
……(3) 
Equation (3) is the required equation. 
(b) 𝑛21 =
𝜆1
𝜆2
=
𝑛2
𝑛1
 
𝑓𝑜𝑟 𝜆2 > 𝜆1 , 𝑛2 𝜆1 , 𝑛2 𝑓1 
Thus, when wavelength of incident light is increased, focal length of the convex lens increases. 
 
66. 
 
 
 
 
 
 
Previous Year CBSE Questions Page 275 of 484 
 
Fig(a) shows the geometry of image formation by a double convex lens. The image formation 
can be seen in terms of two steps: 
(i) The first refracting surface forms the image I1 of the object O (fig b) 
(ii) The image I1 acts as a virtual object for the second surface that forms the image at I (fig 
c) 
Applying equation for refraction at a spherical surface to the first surface ABC, 
𝑛1
𝑂𝐵
+
𝑛2
𝐵𝐼1
=
𝑛2 − 𝑛1
𝐵𝐶1
……(1) 
Applying equation for refraction at a spherical surface to the second surface ADC, 
𝑛2
−𝐷𝐼1
+
𝑛1
𝐷𝐼
=
𝑛1 − 𝑛2
−𝐷𝐶2
…… (2) 
For this case the refractive index of the medium on the right side of ADC is 𝑛1 while on its left it 
is 𝑛2. Further 𝐷𝐼1 is negative as the distance is measured against the direction of incident light. 
Also for thin lens, 𝐵𝐼1 = 𝐷𝐼1 
Equation (2) becomes 
−
𝑛2
𝐵𝐼1
+
𝑛1
𝐷𝐼
=
𝑛2 − 𝑛1
𝐷𝐶2
……(3) 
Adding eqs (1) and (3) 
𝑛1
𝑂𝐵
+
𝑛1
𝐷𝐼
= (𝑛2 − 𝑛1) (
1
𝐵𝐶1
+
1
𝐷𝐶2
)…… (4) 
Suppose the object is at infinity, 𝑂𝐵 → ∞ 𝑎𝑛𝑑 𝐷𝐼 = 𝑓 
𝑛1
𝑓
= (𝑛2 − 𝑛1) (
1
𝐵𝐶1
+
1
𝐷𝐶2
)…… (5) 
By the sign conventions, 𝐵𝐶1 = +𝑅1𝐷𝐶2 = −𝑅2 
𝑛1
𝑓
= (𝑛2 − 𝑛1) (
1
𝑅1
+
1
−𝑅2
) 
1
𝑓
= (
𝑛2
𝑛1
−
𝑛1
𝑛1
) (
1
𝑅1
−
1
𝑅2
)……… (𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 𝑛1) 
1
𝑓
= (
𝑛2
𝑛1
− 1) (
1
𝑅1
−
1
𝑅2
) 
1
𝑓
= (𝑛21 − 1) (
1
𝑅1
−
1
𝑅2
)…… (6) (∵ 𝑛21 =
𝑛2
𝑛1
) 
Eqn (6) is Lens Maker’s formula. 
Previous Year CBSE Questions Page 276 of 484 
 
Assumptions made in the derivation of Lens Maker’s formula 
1. The lens used is thin so that the distances measured from its surfaces may be taken equal 
to those measured from its optical centre. 
2. The object is a point object placed on the principal 
axis. 
3. The aperture of the lens is small. 
4. All the rays are paraxial. 
New Cartesian sign convention for spherical lenses: 
1. All distances are measured from the optical centre of 
the lens. 
2. The distances measured in the direction of incident 
light are positive. 
3. The distances measured in the opposite direction of incident light are negative. 
67. Figure shows the passage of light through a triangular prism ABC. Ray PQ is incident on the 
face AB. The angle of incidence and angle of refraction are i and r1respectively. The ray QR 
emerges from the face AC. The angle of incidence on this face is r2 and angle of refraction or 
emergence is e. 
The angle between the emergent ray RS and the direction of the incident ray is called the angle of 
deviation, δ 
In □ AQNR, ∠𝐴𝑄𝑁 + ∠𝐴𝑅𝑁 = 90𝑂 + 90𝑂 = 180𝑂 
∠𝐴 + ∠𝑄𝑁𝑅 = 180𝑂 … . . (1) 
In ∆QNR, 𝑟1 + 𝑟2 + ∠𝑄𝑁𝑅 = 180𝑂 …… (2) 
Comparing (1) and (2) 𝑟1 + 𝑟2 = 𝐴…… . . (3) 
In ∆MQR, 𝛿 = ∠𝑀𝑄𝑅 + ∠𝑀𝑅𝑄 … . . (𝛿 𝑖𝑠 𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒) 
𝛿 = (𝑖 − 𝑟1) + (𝑒 − 𝑟2) 
𝛿 = 𝑖 + 𝑒 − 𝐴…… . . . (4) 
Plot of angle of deviation (𝜹) versus angle of incidence (i) for a triangular prism: 
Previous Year CBSE Questions Page 277 of 484 
 
 
At 𝛿 = 𝛿𝑚𝑖 = 𝑒 ∴ 𝑟1 = 𝑟2 
∴ 𝑒𝑞𝑛 (3)𝑏𝑒𝑐𝑜𝑚𝑒𝑠 2𝑟 = 𝐴 ∴ 𝑟 =
𝐴
2
 
𝐸𝑞𝑛 (4) 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝛿𝑚 = 2𝑖 − 𝐴 ∴ 𝑖
=
𝐴 + 𝛿𝑚
2𝛿
 
According to Snell’s law 
𝑛21 =
sin 𝑖
sin 𝑟
 
Substituting i and r in the above equation 𝑛21 =
sin(
𝐴+𝛿𝑚
2
)
sin(
𝐴
2
)
 
 
 
************ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Previous Year CBSE Questions Page 278 of 484 
 
SECTION B 
1 MARK QUESTIONS 
Q1. What happens to the focal length of a lens when it is immersed in water? 
Q2. What happens to the power of a lens immersed in water? 
Q3. How can a convex lens behave like a diverging lens? 
Q4. An object is placed at the focus of a concave lens, where will be the image formed? 
Q5. Two lenses having focal lengths 𝑓1𝑎𝑛𝑑 𝑓2 are placed in contact. what is the focal 
length of the combination? 
Q6. A converging and the diverging lens of equal focal lengths are placed co-axially in 
contact. Find the focal length and power of the combination? 
Q7. A double convex lens made from a material of refractive index 𝑛2is immersed in a 
liquid of refractive index𝜇1(𝑛2 > 𝑛1 ) what change if any wouldoccur in the nature 
of the lens? 
Q8. A glass lens of refractive index 1.45 is placed in a liquid, what must be the refractive 
index of the liquid in order to make the lens disappear? 
Q9. For the same value of angle of incidence, the angles of refraction in three media are 
15°, 25° and 35° respectively. In which medium would the velocity of light be 
minimum? 
Q10. A biconvex lens made of a transparent medium of refractive index 1.5 is immersed 
in water of refractive index 1.33. Will the lens behave as a converging or diverging 
lens. Give reason? 
Q11. Draw the path of ray of light suffering minimum deviation while passing through a 
prism? 
Q12. Plot a graph to show the variation of angle of deviation as a function of angle of 
incidence for light rays passing through a prism? 
Q13. Write down the relation between the refractive index of the material of the prism, 
angle of prism and angle of minimum deviation? 
Q14. What do you mean by angle of minimum deviation? 
Q15. What is the relation between refractive index and wavelength of light? 
Q16. Write down the expression for magnifying power of a compound microscope? 
 
Previous Year CBSE Questions Page 279 of 484 
 
MCQ 
Q17. A convex lens of focal length 10 cm is placed in contact with a concave lens of focal 
length 20cm. What is the nature and focal length of the combination? 
a) Concave,10cm 
b) Convex, 10cm 
c) Concave, 20cm 
d) Convex, 20cm 
Q18. A convex lens of focal length 16cm forms a virtual image of double the size of the 
object. What is the distance of the object from the lens? 
a) 8cm b) 16cm c) 24cm d) 32cm 
Q19. The angle of prism is 30º and ray incident at 60º on one refracting surface suffers a 
deviation of 30 º. What is the angle of emergence? 
a) 0 º b) 15 c) 30 º d) 45 º 
Q20. A ray of light passes through an equilateral prism such that the angle of incidence is 
equal to angle of emergence and the later is equal 3/4th the angle of prism. The angle 
of deviation is 
a) 45 º b) 39 º c) 20 º d) 30 º 
Q21. A convex length of focal length 15cm is made of material having refractive index 
1.2. When placed in water of refractive index 1.3 it will behave as 
a) Converging lens of focal length 15cm 
b) Converging lens of focal length different than 15 cm 
c) Diverging lens of focal length 15cm 
d) Diverging lens of focal length different than 15 cm 
 
 
ASSERTION AND REASON 
Q22. Assertion (A): We cannot get diffraction pattern from a wide slit illuminated by 
monochromatic light. 
Reason (R): In diffraction pattern, all the bright bands are not of the same intensity. 
Q23. Assertion (A): When a light wave travels from a rarer to a denser medium, it loses speed. 
The reduction in speed imply a reduction in energy carried by the light wave. 
Reason (R): The energy of a wave is proportional to velocity of wave. 
Q24. Assertion (A): The film which appears bright in reflected system will appear dark in the 
transmitted light and vice-versa. 
Reason (R): The conditions for film to appear bright or dark in reflected light are just reverse 
to those in the transmitted light. 
Previous Year CBSE Questions Page 280 of 484 
 
Q25. Assertion (A): In Young’s double slit experiment, the fringes become indistinct if one of 
the slits is covered with cellophane paper. 
Reason (R): The cellophane paper decrease the wavelength of light. 
Q26. Assertion (A): One of the conditions for interference is that the two sources should be 
very narrow. 
Reason (R): One broad source is equal to large number of narrow sources. 
2 MARKS QUESTION 
Q27. Give reasons to explain why a reflecting telescope is preferred over a refracting 
telescope? 
Q28 Draw a ray diagram to show as to how a right isosceles prism made of crown glass can 
be used to obtain an inverted image? 
Q29 A ray of light incident on a equilateral prism propagates parallel to the baseline of the 
prism inside it. Find the angle of incidence of this ray, given the refractive index of the 
material of the prism is √3 ? 
Q30. What is total internal reflection and what are the conditions under which it occurs? 
Q31. The objective of a telescope is of larger focal length and of larger aperture compared to 
the eyepiece. Give reasons? 
3 MARKS QUESTION 
Q32.(a) With the help of a ray diagram, show how a concave mirror is used to obtain an erect and 
magnified image of an object. 
(b) Using the above ray diagram, obtain the mirror formula and the expression for linear 
magnification. (CBSE 18C, 19C) 
Q33. Draw the ray diagram showing refraction of light through a glass prism and hence obtain the 
relation between the refractive index n of the prism, angle of prism and angle of minimum 
deviation. (CBSE F 17) OR 
(i)Trace the path of a ray of light PQ which is incident at an angle i on one face of a glass prism of 
angle A. It then emerges out from the other face at an angle e. Use the ray diagram to prove that 
the angle through which the ray is deviated is given by ∠𝛿 = ∠𝑖 + ∠𝑒 − ∠𝐴 
(ii) What will be the minimum value of δ if the ray passes symmetrically? (CBSE F 2022) 
Q34. (a) Draw a ray diagram showing the image formation by a compound microscope. 
(b) Derive an expression for total magnification when the image is formed at infinity.