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Hint: To provide angular magnification of distant objects, the telescope is used. As the object is very far away, the objective has a large focal length and a much larger aperture than the eyepiece. To solve this question, we must know the formula for magnification in telescopes.
Formula Used: We will use the following formula to solve this question
$m = \dfrac{{{\alpha _i}}}{{{\alpha _o}}} = \dfrac{{{f_o}}}{{{f_e}}}$
Where
$m$ is the magnification
${\alpha _o}$ is the angle subtended by the object
${\alpha _i}$ is the angle subtended by the image
${f_o}$ is the focal length of objective
${f_e}$ is the focal length of the eyepiece
Complete Step-by-Step Solution:
Light enters the objective from a distant object and at its second focal point, a real and inverted image is formed.
This image acts as an eyepiece object; it magnifies this image and generates a final inverted image.
The following information is provided to us in the question
The focal length of objective, ${f_o} = 60 cm$
The focal length of the eyepiece, ${f_e} = 5 cm$
The angle subtended by the object, ${\alpha _o} = {2^ \circ }$
Now, let us use the formula given above
$m = \dfrac{{{\alpha _i}}}{{{\alpha _o}}} = \dfrac{{{f_o}}}{{{f_e}}}$
Let us now substitute the known values in the formula to get the required result
\[\dfrac{{{\alpha _i}}}{2} = \dfrac{{60}}{5}\]
Upon solving further, we get
\[\therefore {\alpha _i} = {24^ \circ }\]
Hence, the correct option is (B.)
Note: Real and highly diminished images are formed by a telescope. When an object is placed far from the focal length of a convergent lens, depending on how close the object is to the focal point, the lens produces an inverted, real and enlarged image. Telescopes are visual instruments used in the field of astronomy that use lenses and mirrors. There are many other classifications of telescopes according to their varying tasks.
Formula Used: We will use the following formula to solve this question
$m = \dfrac{{{\alpha _i}}}{{{\alpha _o}}} = \dfrac{{{f_o}}}{{{f_e}}}$
Where
$m$ is the magnification
${\alpha _o}$ is the angle subtended by the object
${\alpha _i}$ is the angle subtended by the image
${f_o}$ is the focal length of objective
${f_e}$ is the focal length of the eyepiece
Complete Step-by-Step Solution:
Light enters the objective from a distant object and at its second focal point, a real and inverted image is formed.
This image acts as an eyepiece object; it magnifies this image and generates a final inverted image.
The following information is provided to us in the question
The focal length of objective, ${f_o} = 60 cm$
The focal length of the eyepiece, ${f_e} = 5 cm$
The angle subtended by the object, ${\alpha _o} = {2^ \circ }$
Now, let us use the formula given above
$m = \dfrac{{{\alpha _i}}}{{{\alpha _o}}} = \dfrac{{{f_o}}}{{{f_e}}}$
Let us now substitute the known values in the formula to get the required result
\[\dfrac{{{\alpha _i}}}{2} = \dfrac{{60}}{5}\]
Upon solving further, we get
\[\therefore {\alpha _i} = {24^ \circ }\]
Hence, the correct option is (B.)
Note: Real and highly diminished images are formed by a telescope. When an object is placed far from the focal length of a convergent lens, depending on how close the object is to the focal point, the lens produces an inverted, real and enlarged image. Telescopes are visual instruments used in the field of astronomy that use lenses and mirrors. There are many other classifications of telescopes according to their varying tasks.
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