A terrestrial telescope has an objective lens and eyepiece of focal length $200 cm$ and $4 cm$ respectively. Separation between the objective lens and the eyepiece is $220 cm$ . The power of the erecting lens is $n^{2} D$ . The value of $n$ (integer only)

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Solution

Given,

$f_{0} = 200 cm; f_{e} = 4 cm; and L = 220 cm$

Where, $f_{0} = Focal length of the objective f_{e} = Focal length of the eyepiece$

For normal adjustment, the length of the telescope is,

$L = f_{0} + 4 f + f_{e}$

Where, $f$ is the focal length of the erecting lens.

$220 = 200 + 4 f + 4$

$\Rightarrow 4 f = 220 - 204 = 16$

$∴ f = 4 cm$

Now, the power of the erecting lens is,

$P = \frac{1}{f} = \frac{100}{4} m = 25 D$

$\Rightarrow P = 5^{2} D$

Given, $P = n^{2} D$

$∴ n = 5$

Key Note:
$(i)$ A terrestrial telescope is used to see object at long distance on the surface of earth. Hence, image should be erect.
$(i i)$ A terrestrial telescope has an additional erecting lens is used to make the final image erect.

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A terrestrial telescope has an objective lens and eyepiece of focal length 200 cm and 4 cm respectively. Separation between the objective lens and the eyepiece is 220 cm. The power of the erecting lens is n2 D. The value of n(integer only)

A terrestrial telescope has an objective lens and eyepiece of focal length $200 cm$ and $4 cm$ respectively. Separation between the objective lens and the eyepiece is $220 cm$ . The power of the erecting lens is $n^{2} D$ . The value of $n$ (integer only)