SATHEE: ray-optics-and-optical-instruments Question 43

ray-optics-and-optical-instruments Question 43

Question: Q. 4. (i) A giant refracting telescope has an objective lens of focal length $15 m$ . If an eye piece of focal length $1.0 cm$ is used, what is the angular magnification of the telescope?

(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^{6} m$ and the radius of lunar orbit is $3.8 \times 10^{8} m$ .

A [Delhi I, II, III 2015]

Detailed Answer :

(i) $m = \frac{- f_{0}}{f_{e}}$ ; ignoring -ve sign as it only shown that image is inverted.

$\begin{aligned} f_{o} = 1500 cm & f_{e} = 1 cm & m = \frac{1500}{1} = 1500 \end{aligned}$

(ii) Angular size of the moon,

$\tan α = \frac{h_{0}}{u_{0}}$

Angular size of the moon’s image by objective lens is also, $\tan α = \frac{h_{1}}{f_{o}}$

Hence,

$\begin{aligned} \frac{h_{0}}{u_{0}} & = \frac{h_{1}}{f_{0}} h_{o} & = 3.48 \times 10^{6} m u_{o} & = 3.8 \times 10^{8} m . f_{o} & = 15 m \frac{3.48 \times 10^{6}}{3.8 \times 10^{8}} & = \frac{h_{1}}{15} \end{aligned}$

$\begin{aligned} h_{1} & = \frac{3.48 \times 10^{6}}{3.8 \times 10^{8}} \times 15 & = 13.7 cm \end{aligned}$

Answering Tip

Learn all the formulae of Telescope carefully. Do practice to solve the humericals by taking care of their sign convention.

Q. 5. (i) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at the least distance of distinct vision.

(ii) The total magnification produced by a compound microscope is 20 . The magnification produced by the eye piece is 5 . The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be $14 cm$ .

If the least distance of distinct vision is $20 cm$ , calculate the focal length of the objective lens and the eye piece.

A [Delhi I, II, III 2014]

Show Answer

Solution:

Ans. (i) Labelled ray diagram of a compound microscope for formation of image at the near point of the eye :

(ii) Given, $m = 20, m_{e} = 5, D = 20 cm$

For eyepiece, $m_{e} = \frac{D}{u_{e}}$

$5 = - \frac{20}{u_{e}}$

$\begin{aligned} u_{e} & = - 4 cm \frac{1}{f_{e}} & = \frac{1}{v_{e}} - \frac{1}{u_{e}} & = \frac{1}{- 20} + \frac{1}{4} \frac{1}{f_{e}} & = \frac{- 1 + 5}{20} \frac{1}{f_{e}} & = \frac{4}{20} f_{e} & = 5 cm \end{aligned}$

Hence, focal length of eyepiece, $f_{e} = 5 cm$

$\begin{aligned} m & = m_{o} \times m_{e} m_{o} & = \frac{m}{m_{e}} = \frac{20}{5} = 4 v_{o} & = 14 - 4 = 10 cm \end{aligned}$

For objective lens,

$\begin{aligned} m_{o} & = - \frac{v_{0}}{u_{o}} 4 & = - \frac{10}{u_{o}} (∵ m_{o} = 4) \Rightarrow u_{o} & = - \frac{10}{4} = - 2.5 cm Now, \frac{1}{f_{0}} & = \frac{1}{v_{o}} - \frac{1}{u_{0}} \frac{1}{f_{0}} & = \frac{1}{10} - \frac{1}{- 2.5} \frac{1}{f_{0}} & = \frac{1}{10} + \frac{10}{25} \frac{1}{f_{0}} & = \frac{1}{2} f_{0} & = 2 cm \end{aligned}$

Hence, focal length of objective, $f_{0} = 2.0 cm$ .

Ray Optics And Optical Instruments

ray-optics-and-optical-instruments Question 43

Detailed Answer :

Answering Tip

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Ray Optics And Optical Instruments

ray-optics-and-optical-instruments Question 43

Detailed Answer :

Answering Tip

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu