Name: Total Internal Reflection From Prism
Uploaded: 2019-07-19
Duration: 1 min 56 s
Description: You must have seen reflectors on a bicycle. Why are they attached to the mudguards? What use do they have? Let’s watch this video on reflectors to find out.

Question

Draw the ray diagram to show the formation of the image by compound microscope and obtain an expression for the magnifying power when the final image is formed at least distance of distinct vision.

Accepted Answer

Compound microscope consists of two convex lenses separated by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of larger focal length. The object placed near the objective lens produces an image and this image acts as object for the eye piece lens. The position of the eye piece lens is adjusted such that it produces virtual and magnified image of the object.

The magnifying power of a microscope is the ratio of the angle subtended by image to the angle subtended by image placed at least distance of distinct vision.
Therefore, from the above diagram,
$m = \frac{β}{α}$ ….. (1)
$∴ α = \frac{A B}{D}$ …… (2)
Where $D$ is the least distance of distinct vision i.e. the distance between point O and Point B
Now, the angle made by image $A' B'$ ' can be given by,

$β = \frac{A' B'}{E B'}$

Let, $u_{e}$ be the distance between image and eyes piece
Thus, $β = \frac{A' B'}{u_{e}}$ ……. (3)
Now, from (1), (2) and (3)
We get,

$m = \frac{A' B'}{u_{e}} \times \frac{D}{A B}$

From the diagram above we can say that,

$\frac{A' B'}{A B} = \frac{O B'}{O B}$

Substitute this in the above equation,

$m = \frac{O B'}{O B} \times \frac{D}{u_{e}}$

Here, $O B$ and $O B ’$ is the distance of object and image from point $O$ .
Therefore,
$m = \frac{v_{0}}{u_{0}} \times \frac{- D}{- u_{e}}$ ……. (4)
On solving,
$m = \frac{v_{0} D}{u_{0} u_{e}}$

Using the lens formula, we know,

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

The lens formula for the microscope:

$\frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}$

Since the distance $D$ is positive.
Therefore,

$\frac{1}{- f_{e}} = \frac{1}{D} - \frac{1}{u_{e}}$

Rearranging the terms, We get,
$\frac{1}{u_{e}} = \frac{1}{f_{e}} + \frac{1}{D}$

On solving,

$\frac{D}{u_{e}} = \frac{D}{f_{e}} + 1$ ….. (5)
Now, from (4) and (5), We get,

$m = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}})$

Therefore, the magnification power of a compound microscope is given by, $m = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}})$

Draw the ray diagram to show the formation of the image by compound microscope and obtain an expression for the magnifying power when the final image is formed at least distance of distinct vision.

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