First slide

Ray optics

Question

A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance

Difficult

A

37.3cm
B

46.0cm
C

50.0cm
D

54.0cm

Solution

$f_{o}$ = 40cm ; $f_{e}$ = 4cm ; u₀ = 200cm
$\frac{1}{f_{0}} = \frac{1}{v_{0}} - \frac{1}{u_{0}}; V_{0} = 50 cm; L = V_{0} + f_{e}$
= 50 + 4 = 54cm
when image is formed at least distance of distinct vision
$\frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}; L_{D} = V_{0} + u_{e} V_{e} = - 25 cm \frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}; u_{e} = \frac{100}{29} L_{D} = 50 + \frac{100}{29} = 53.44 cm$

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