Optics Question and Answer

Question 1:
Figure 1: FBD Sketch
Figure 2: Lens Sketch
D
2 πd = A
360 , whereA= Angular
h
2 π ×17 = A
360 ,henceA = h
17 × 360
2 π
From the given data,
f = 8cm
u = 7cm
From lens formula,
1
v + 1
u = 1
f
OR

1
v = 1
f −1
u
Substituting the values of f and u:
1
v = 1
8 − 1
7 =−1
56
Hence, v = -56 cm
Magnification:
M =−v
u = h '
h =−−56
7 = h '
h = 8
1 =8 ,where:
h = height of the object, and
h’ = height of the image
h '
2 πd = AL
360 ° ,
d = (v+10) = 66 cm
Hence, h'
2 π ×66 = AL
360 ° => AL= h '
66 × 360 °
2 π

Question 2
0.01= θ−sinθ
θ
1− sinθ
θ =0.01
sinθ
θ =0.99
Now for
θ=0.245 rad
sinθ
θ =0.99
Hence
θ=0.245 rad
Question 3
(a) Using small angle approximation,
The moon is located at =384400km from the earth. Its radius is 3474/2=1737. Its
angular radius from the earth =1737/384400=0.004519
Because 1 radian =57.3 degrees, the angular radius of the moon
=0.004519*57.3=0.2589degress.
0.2589degress=15.534 minutes of an arc
(b) From Rayleigh Criterion
θ=1.22 λ
b = 1.22× 500 ×10−9
25.08× 103 =2.4322 ×10−13
Since the darkness-adapted pupil is 7mm, the resolution is adequate to see the crater.
(c) The lens has a larger aperture. The larger the aperture of a lens the better the
resolution of the lens. It will thus be possible to view the crater using the lens.