Name: Ray Diagram: Lenses
Uploaded: 2019-07-19
Duration: 6 min 35 s
Description: In this video, let us discuss how ray diagrams for lenses can be used to determine the size and location of a refracted image. Get in to know more!

Question

With the help of labelled ray diagram, explain the formation of final image which is formed at least distance of distinct vision in case of astronomical telescope. Define its magnifying power and derive an expression for its magnifying power.

Accepted Answer

Astronomical telescope:

Image formation: A parallel beam of light from a heavenly body such as stars, planets, or satellites falls on the objective lens of a telescope. The objective lens forms a real, inverted, and diminished image $A^{'} B^{'}$ of the heavenly body. This image now acts an object for the eyepiece $E$ , whose position is adjusted so that the image lies between the focus $f_{e}$ and optical centre $C_{2}$ of the eyepiece. Now the eyepiece forms a virtual, inverted and highly magnified image of object at infinity or least distance of distinct vision. When the final image of an object is formed at infinity or least distance of distinct vision, the telescope is said to be in normal adjustment. The final image of an object formed by an astronomical telescope is always inverted with respect to the object.

Magnifying power: Magnifying power of an astronomical telescope is defined as the ratio of the angle subtended at the eye by the final image as seen through the telescope to the angle subtended at the eye by the object seen directly when both and object are at infinity.

Expression for magnifying power:

From the figure

$\tan (α^{'}) \approx α^{'} = α = \frac{P^{'} Q^{'}}{O P^{'}} = \frac{P^{'} Q^{'}}{f_{o}} . . . . (i)$

Here, $f_{o}$ is the focal length of an objective lens.

The final image $P^{''} Q^{''}$ subtends an angle $β$ on the eyepiece.

From the $△ P^{'} Q^{'} E$

$β \approx \tan (β) = \frac{P^{'} Q^{'}}{E P^{'}} . . . . (ii)$

From the above equation

$\frac{β}{α} = \frac{f_{o}}{E^{'} P^{'}}$

If the telescope is set for normal adjustment so that the final image is formed at infinity. The first image $P^{'} Q^{'}$ must be in the focal plane of the eyepiece.

Then $E^{'} P^{'} = f_{e}$

$\frac{β}{α} = \frac{f_{o}}{f_{e}}$

According to the magnifying power definition

$m = \frac{β}{α}$

So, $m = \frac{f_{o}}{f_{e}}$

Since the angles $β$ and $α$ are formed on the opposite sides of the axis.

Therefore,

$m = - \frac{f_{o}}{f_{e}}$

With the help of labelled ray diagram, explain the formation of final image which is formed at least distance of distinct vision in case of astronomical telescope. Define its magnifying power and derive an expression for its magnifying power.

Important Questions on Ray Optics and Optical Instruments

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