Question

A telescope consists of two thin lenses of focal lengths \[{\text{0}}{\text{.3m}}\] and \[{\text{3cm}}\] respectively. It is focused on moon which subtends an angle of \[{\text{0}}{\text{.}}{{\text{5}}^{\text{o}}}\] at the objective. Then, the angle subtended at the eye by the final image will be:
(A) \[{{\text{5}}^{\text{o}}}\]
(B) \[{\text{0}}{\text{.2}}{{\text{5}}^{\text{o}}}\]
(C) \[{\text{0}}{\text{.}}{{\text{5}}^{\text{o}}}\]
(D) \[{\text{0}}{\text{.3}}{{\text{5}}^{\text{o}}}\]

Answer 1

Hint:Angular magnification is equivalent to the number of angle tangents subtended by an object and its reflection, as with magnifiers and binoculars, when determined from a given point of the instrument.
First find the angular magnification of telescope using the values of \[{f_0}\] and \[{f_e}\] then again find the angular momentum focused on moon which subtends an angle of \[{\text{0}}{\text{.}}{{\text{5}}^{\text{o}}}\] at the objective from it.
The formula for angular magnification that we use here is \[m = - \dfrac{{{f_o}}}{{{f_e}}}\]

Complete step by step answer:
Here,
We know that,
Angular magnification of telescope is,
\[m = - \dfrac{{{f_o}}}{{{f_e}}}\]
Given,
\[{f_0} = 0.3{\text{m}}\]
\[{f_e} = 3{\text{cm = }}\dfrac{3}{{100}}{\text{m = 0}}{\text{.03m}}\]
Therefore,
\[m = - \dfrac{{0.3}}{{0.03}} = - 10\]
Also,
Angular magnification is,
\[m = \dfrac{{{\theta _i}}}{{{\theta _o}}}\]
Given,
\[m = - 10\]
\[{\theta _o} = {0.5^{\text{o}}}\]
Therefore,
$
   - 10 = \dfrac{{{\theta _i}}}{{{{0.5}^{\text{o}}}}} \\
   \Rightarrow {\theta _i} = - {5^{\text{o}}} \\
$
Here, \[{\theta _i}\] shows a negative value which denotes an inverted image.
Hence, the angle subtended at the eye by the final image will be \[{5^{\text{o}}}\].
So, the correct answer is (A).

Note: When measured from a given point in the instrument, angular magnification is equal to the ratio of the tangents of the angles subtended by an object and its image, as with magnifiers and binoculars. An optical system's focal length is a measure of how intensely the system converges or diverges light; it's the opposite of the optical strength of the system. A positive focal length means that light is converged by a system, while a negative focal length implies light is diverged by the system.