Question

A simple telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cm is focussed on a distant object in such a way that parallel rays emerge from the eye lens. If the object subtends an angle of ${{2}^{\circ }}$ at the objective, the angular width of the image is
A. ${{10}^{\circ }}$
B. \[{{24}^{\circ }}\]
C. ${{50}^{\circ }}$
D. ${{\left( \dfrac{1}{6} \right)}^{\circ }}$

Answer 1

Hint: Magnification of telescope is the ratio of focal length of eyepiece to the focal length of objective and magnification can be defined as the angle subtended by object to the angle subtended by image.

Complete answer:
Given,
Basically a simple telescope contains a pair of convex lenses. These lenses are mounted in a tube.
The lens of lesser focal length is known as an eyepiece lens and the lens of greater focal length is known as an objective lens.
Focal length of simple telescope is 60 cm
Focal length of single eye lens of focal length is 5 cm
Angle subtended by object = ${{2}^{\circ }}$
Magnification of telescope is given as

$M=\dfrac{focal\,length\,of\,eye\,piece}{focal\,length\,of\,objective}=\dfrac{angle\,subtended\,by\,object}{angle\,subtended\,by\,image}$
\[\Rightarrow \dfrac{5}{60}=\dfrac{{{2}^{\circ }}}{angle\,subtended\,by\,image}\]
$\Rightarrow angle\,subtended\,by\,image\,={{24}^{\circ }}$
So, the correct answer is “Option B”,${{24}^{\circ }}$

Note:
In the given question both the value of magnification equated then we got the angle subtended by the image which is also known as the angular width of the image. Magnification of a telescope is the ratio of focal length of eyepiece to the focal length of objective and magnification can be defined as the angle subtended by object to the angle subtended by image.