Question: Q. 2. Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain.

U [Delhi I, 2017]

Show Answer

Solution:

Ans. Magnifying power is defined as the angle subtended by the image to the angle subtended (at the unaided eye) by the object.

(Alternatively : Also accept this definition in the form of formula)

$$ m=m_{0} \times m_{e}=\frac{L}{f_{0}} \times \frac{D}{f_{e}} $$

To increase the magnifying power both the objective and eyepiece must have short focal lengths $\left(\right.$ as $\left.m=\frac{L}{f_{o}} \times \frac{D}{f_{e}}\right)$ $1 / 2+1 / 2$

[CBSE Marking Scheme 2017]

Commonly Made Error

  • Students wrote correct definition of magnifying power of the microscope but wrote wrong formula.

[AI Q. 3. (i) State the condition under which a large magnification can be achieved in an astronomical telescope.

(ii) Give two reasons to explain why a reflecting telescope is preferred over a refracting telescope.

U[CBSE 2017, Foreign set]

Ans. (i)

$$ m=\frac{f_{0}}{f_{e}} $$

By increasing $f_{o}$ or decreasing $f_{e}$

(ii) (a) No chromatic aberration.

(b) No spherical aberration.

(c) Mechanical advantage - low weight, easier to support.

(d) Mirrors are easy to prepare.

(e) More economical.

(Any two) $1 / 2+1 / 2$

[CBSE Marking Scheme 2017] AT Q. 4. You are given two converging lens of focal lengths $1.25 \mathrm{~cm}$ and $5 \mathrm{~cm}$ to design a compound microscope. If it is desired to have a magnification of 30, find out the separation between the objective and the eyepiece.

A [OD 2015]

Ans. Given, $f_{o}=1.25 \mathrm{~cm}, f_{e}=5 \mathrm{~cm}$

Magnification, $\quad m=30$,

If we set these lens for minimum distance for distinct vision, then for

$$ \begin{align*} m & =\frac{L}{f_{o}}\left(1+\frac{D}{f_{e}}\right) \ m & =30 \ \mathrm{~L} & =6.25 \mathrm{~cm} \tag{1} \end{align*} $$

Hence, distance between two lenses is

$$ \begin{aligned} & =f_{o}+6.25+f_{e} \ & =(1.25+6.25+5.0) \mathrm{cm} \ & =12.5 \mathrm{~cm} \end{aligned} $$

This is a required separation between the objective and the eyepiece. [AI Q. 1. Draw a labelled ray diagram to show the image formation in a refracting type astronomical telescope in the normal adjustment position. Write two drawbacks of refracting type telescopes.

A [CBSE SQP 2018-19]

Drawbacks :

(i) Large sized lenses area heavy and difficult to support.

(ii) Large sized lenses suffer from chromatic and spherical aberration.

[CBSE Marking Scheme 2018]



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