Question

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens. Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.

(i) The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will be (a) 3.45 cm (b) 5 cm (c) 1.29 cm (d) 2.59 cm

(ii) How far from the objective should an object be placed in order to obtain the condition described in part (i) (a) 4.5 cm (b) 2.5 cm (c) 1.5 cm (d) 3.0 cm

(iii) What is the magnifying power of the microscope in case of least distinct vision? (a) 20 (b) 30 (c) 40 (d) 50

(iv) The magnifying power of a compound microscope increases with (a) the focal length of objective lens is increased and that of eye lens is decreased. (b) the focal length of eye lens is increased and that of objective lens is decreased. (c) focal length of both objects and eye-piece are increased. (d) focal length of both objects and eye-piece are decreased.​

Answer 1

Concept Introduction:-

The most typical kinds of microscopes which consists of two parts a type of optical instrument that allows you to see highly magnified images of small things are referred to compound microscope.

Given Information:-

We have been given that  a compound microscope that consists of an objective lens of focal length $2.0$ cm and an eyepiece of focal length $6.25$ cm separated by a distance of $15$ cm.

To Find:-

We have to find that The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will be, How far from the objective should an object be placed in order to obtain the condition described in part $(i)$, What is the magnifying power of the microscope in case of least distinct vision and The magnifying power of a compound microscope increases with.

Solution:-

According to the problem

Focal length of the objective lens $\mathrm{f}_{1}=2 \mathrm{~cm}$

Focal length of the eyepiece, $f_{2}=6.25 \mathrm{~cm}$

Distance between the objective lens and the eyepiece, $\mathrm{d}=15 \mathrm{~cm}$

Least distance of distinct vision, $\mathrm{d}^{\prime}=25 \mathrm{~cm}$

Image distance for the eyepiece, $\mathrm{v}_{2}=-25 \mathrm{~cm}$

object distance for the eyepiece $=u_{2}$

According to the lens formula

$\begin{aligned}&\frac{1}{\mathrm{f}_{2}}=\frac{1}{\mathrm{v}_{2}}-\frac{1}{\mathrm{u}_{2}} \\&\frac{1}{6.25}=\frac{1}{-25}-\frac{1}{\mathrm{u}_{2}} \\&\mathrm{u}_{2}=-5 \mathrm{~cm}\end{aligned}$

Image distance for the objective lens

$\mathrm{v}_{1}=\mathrm{d}+\mathrm{u}_{2}=15-5=10 \mathrm{~cm}$

According to the lens formula

$\begin{aligned}&\frac{1}{\mathrm{f}_{1}}=\frac{1}{\mathrm{v}_{1}}-\frac{1}{\mathrm{u}_{1}} \\&\frac{1}{2}=\frac{1}{10}-\frac{1}{\mathrm{u}_{1}} \\&\mathrm{u}_{2}=-2.5 \mathrm{~cm}\end{aligned}$

The magnifying power of a compound microscope is given by the relation

$\begin{aligned}&\mathrm{m}=\frac{\mathrm{v}_{1}}{\left|\mathrm{u}_{1}\right|}\left(1+\frac{\mathrm{d}^{\prime}}{\mathrm{f}_{2}}\right) \\&\mathrm{m}=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=20\end{aligned}$

The final image is formed at infinity

Image distance for the eyepiece $v_{2}=\infty$

According to the lens formula

$\frac{1}{\mathrm{f}_{2}}=\frac{1}{\mathrm{v}_{2}}-\frac{1}{\mathrm{u}_{2}}$

According to the lens formula

$\frac{1}{6.25}=\frac{1}{\infty}-\frac{1}{\mathrm{u}_{2}}$

$\mathrm{u}_{2}=-6.25 \mathrm{~cm}$

Image distance for the objective lens

$\mathrm{v}_{1}=\mathrm{d}+\mathrm{u}_{2}=10-6.25=8.75$

Object distance for the objective len $=\mathrm{u}_{2}$

According to the lens formula

$\frac{1}{\mathrm{f}_{1}}=\frac{1}{\mathrm{v}_{1}}-\frac{1}{\mathrm{u}_{1}}\\\frac{1}{2}=\frac{1}{8.75}-\frac{1}{u_{1}}\\\mathrm{u}_{2}=-2.59 \mathrm{~cm}$

The magnifying power of a compound microscope is given by the relation:

$\frac{\mathrm{v}_{1}}{\left|\mathrm{u}_{1}\right|}\left(\frac{\mathrm{d}^{\prime}}{\left|\mathrm{u}_{2}\right|}\right)=13.51$

Final Answer:-

The correct answers are $5$cm, $2.5$ cm, $20$ cm and focal length of both objects and eye-piece are decreased.

#SPJ3

Answer 2

Answer:

........................................