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Question : 35
Total: 37
SECTION - D
(a) Define the term 'focal length of a mirror' with the help of any diagram. Obtain the relation between focal length and radius of curvature.
(b) Calculate the angle of emergence (e) of the ray of light incident normally on the face AC of a glass prism
How will the angle of emergence change qualitatively, if the ray of light emerges from the prism into a liquid of refractive index 1.3 instead of air?
OR
(a) Define the term 'resolving power of a telescope'. How will the resolving power be effected with increase in
(i) Wavelength of light used.
(ii) Diameter of the objective lens.
Justify your answers.
(b) A screen is placed
Solution:
(a) Focal length of mirror : When rays of light parallel to the principal axis of a mirror is incident on it, the rays after reflection, either converge at a point or appear to diverge from a point. The distance of that point from the pole of the mirror is known as the focal length of the mirror.
Relation between focal length and radius of curvature:
A ray of lightBP ′ P ′ R
For concave mirror, it passes through the focus . For convex mirror, extending the ray backward it appears to pass through the focus.
P F
PF = f .
C
PC = radius of curvature = R
P ′ C P ′
For concave mirror,
and∠ BP ′ C = ∠ P ′ CF = θ (alternate angles)
∠ BP ′ C = ∠ CP ′ F = θ
(law of reflection,∠ i = ∠ r
Hence∠ P ′ CF = ∠ CP ′ F
∴ △ FP ′ C
Hence,P ′ F = FC
If the aperture of the mirror is small, the pointP ′ P
thenP ′ F = PF
∴ PF = FC
=
PC
∴ f =
R
(b)∠ C = 60 ∘
∠ B = 90 ∘
∠ A = 30 ∘
∴ AB = 30 ∘
=
Or, s i n e = √ 3 s i n 30 ∘ = √ 3 × 1 ∕ 2 = 0.87
∴ e = s i n − 1 0.87 = 60.46 ∘
Now, the prism is immersed in a liquid of refractive index 1.3.
The refractive index of the surrounding medium is now greater than that of air but less than that of the medium of prism. Now, the angle of emergence be less than60.46 ∘
OR
(a) Resolving power is the ability of the telescope to distinguish clearly between two points whose angular separation is less than the smallest angle that the observer's eye can resolve.
Resolving power of a telescope= R =
where,a = λ =
(i) When wavelength of the light increases, the resolving power of the telescope decreases.
(ii) When diameter of the objective increases, the resolving power of the telescope increase.
(b)
For position 1:
u = − x
v = 80 − x
−
=
For position 2:
u = − ( x + 20 )
v = 80 − ( x + 20 )
−
=
Comparing equations (i) and (ii)
−
=
−
Or,
−
=
−
Or,
=
Or, x ( 80 − x ) = ( x + 20 ) ( − x + 60 )
Or, 80 x = 40 x + 1200
∴ x = 30 cm
Putting the value ofx
−
=
Or, −
=
∴ f = −
= 18.75 cm
Relation between focal length and radius of curvature:
A ray of light
For concave mirror, it passes through the focus . For convex mirror, extending the ray backward it appears to pass through the focus.
For concave mirror,
and
(law of reflection,
Hence
Hence,
If the aperture of the mirror is small, the point
then
(b)
Now, the prism is immersed in a liquid of refractive index 1.3.
The refractive index of the surrounding medium is now greater than that of air but less than that of the medium of prism. Now, the angle of emergence be less than
OR
(a) Resolving power is the ability of the telescope to distinguish clearly between two points whose angular separation is less than the smallest angle that the observer's eye can resolve.
Resolving power of a telescope
where,
(i) When wavelength of the light increases, the resolving power of the telescope decreases.
(ii) When diameter of the objective increases, the resolving power of the telescope increase.
(b)
For position 1:
For position 2:
Comparing equations (i) and (ii)
Putting the value of
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