The magnification produced by a astronomical telescope for normal adjustment is 10 and the length of the telescope is 1.1 m. The magnification, when the image is formed atleast distance of distinct vision is

6

18

16

14

Solution

Given:-

m = 10, length of telescope= 1.1 m

We know that,

Magnification m = $\frac{f_{0}}{f_{e}}$

10 = $\frac{f_{0}}{f_{e}}$

f₀ = 10 f_e

∴ f_e + f₀ = 1.1 m

⇒ f_e + 10 f_e = 1.1 mm ....[ $∵$ f₀ = 10f_e ]

⇒ f_e ( 1 + 10 ) = 1.1 m

⇒ f_e = 0.1 m or 10 cm

Magnification least distance of distinct vision,

M_b = $\frac{f_{0}}{f_{e}} (1 + \frac{f_{e}}{D})$

= 10 $(1 + \frac{10}{25})$ .....[ since D = 25 cm ]

= 10 × $\frac{35}{25}$

M_b = 14

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