Optical Instruments And Some Natural Phenomena(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

OPTICAL INSTRUMENTS AND SOME NATURAL PHENOMENA MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm an object is placed at 2 cm form objective and the final image is formed at 25 cm from eye lens. The distance between the two lenses is

6.00 cm
7.75 cm
9.25 cm
11.00 cm

Discuss Question

Answer: Option D. -> 11.00 cm
:
D
It is given that

f_{0} = 1.5 c m, f_{e} = 6.25 c m, u_{0} = 2 c m

When final image is formed at least distance of distinct vision, length of the tube

L_{D} = \frac{u_{0} f_{0}}{u_{0} - f_{0}} + \frac{f_{e} D}{f_{e} + D}

\Rightarrow L_{D} = \frac{2 \times 1.5}{2 - 1.5} + \frac{6.25 \times 25}{6.25 + 25} = 11 c m .

Question 22. Image is formed for the short sighted person at

[AFMC 1988]

Retina
Before retina
Behind the retina
Image is not formed at

Discuss Question

Answer: Option B. -> Before retina
:
B
In short sightedness, the focal length of eye lens decreases, so image is formed before retina.

Question 23. The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is

30 cm
25 cm
15 cm
12 cm

Discuss Question

Answer: Option C. -> 15 cm
:
C
Given that

f_{0} = 1 c m, f_{e} = 5 c m, m_{\infty} = 45

By using

m_{\infty} = \frac{(L_{\infty} - f_{0} - f_{e})}{f_{0} f_{e}} \Rightarrow 45 = \frac{(L_{\infty} - 1 - 5) \times 25}{1 \times 5} \Rightarrow L_{\infty} = 15 c m

Question 24. If the tube length of astronomical telescope is 105 cm long and magnifying power is 20 for normal setting. Calculate the focal length of objective.

100 cm
10 cm
20 cm
25 cm

Discuss Question

Answer: Option A. -> 100 cm
:
A

m = \frac{f_{0}}{f_{e}}; 20 = \frac{f_{0}}{f_{e}} f_{e} = \frac{f_{0}}{20} L = f_{0} + f_{e} 105 = f_{0} + f_{e} 105 = f_{0} + \frac{f_{0}}{20} 105 = (\frac{20 + 1}{20}) f_{0} f_{0} = \frac{105 \times 20}{21} f_{0} = 100 c m

Question 25. If an object subtend angle of

2^{\circ}

at eye when seen through telescope having objective and eyepiece of focal length

f_{0} = 60 c m

and

f_{e} = 5 c m

respectively then angle subtend by image at eye piece will be

16∘
50∘
24∘
10∘

Discuss Question

Answer: Option C. -> 24∘
:
C
By using

\frac{β}{α} = \frac{f_{0}}{f_{e}} \Rightarrow \frac{β}{2^{\circ}} \Rightarrow β = 24^{\circ}

Question 26. The magnifying power of a microscope with an objective of 5 mm focal length is 400. The length of its tube is 20 cm. Then the focal length of the eye-piece is (consider that final image is formed at infinity)

200 cm
160 cm
2.5 cm
0.1 cm

Discuss Question

Answer: Option C. -> 2.5 cm
:
C

m_{\infty} = \frac{(L - f_{o} - f_{e}) D}{f_{o} f_{e}} ≃ \frac{L D}{f_{o} f_{e}}

\Rightarrow 400 = \frac{20 \times 25}{0.5 \times f_{e}} \Rightarrow f_{e} = 2.5 c m .

Question 27. In a compound microscope, the focal length of the objective and the eye lens are 2.5 cm and 5 cm respectively. An object is placed at 3.75 cm before the objective and image is formed at the least distance of distinct vision, then the distance between two lenses will be (i.e. length of the microscopic tube)

11.67 cm
12.67 cm
13.00 cm
12.00 cm

Discuss Question

Answer: Option A. -> 11.67 cm
:
A

L_{D} = v_{o} + u_{e}

and for objective lens

\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}

Putting the values with proper sigh convention.

\frac{1}{+ 2.5} = \frac{1}{v_{o}} - \frac{1}{(- 3.57)} \Rightarrow v_{o} = 7.5_{c m}

For eye lens

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

\Rightarrow \frac{1}{+ 5} = \frac{1}{(- 25)} - \frac{1}{u_{e}} \Rightarrow u_{e} = - 4. 16 c m

\Rightarrow | u_{e} | = 4.16 c m

Hence

L_{D} = 7.5 + 4.16 = 11.67_{c m}

Question 28. A man who cannot see clearly beyond 5 m wants to see stars clearly. He should use a lens of focal length
[MP PET/PMT 1988; Pb. PET 2003]

- 100 m
+ 5 m
- 5 m
Very large

Discuss Question

Answer: Option C. -> - 5 m
:
C
Far point of the myopiceye is the focal length. So, according to sign convention, f= -5 cm.

Question 29. For astronomical telescope
i) focal length of its objective should be larger than that of eyepiece
ii) focal length of its eyepiece should be larger than that of objective
iii) aperture of objective should be wider than that of eye piece
iv) aperture of the objective should be less wider than that of eye piece

i, ii. iii are true
ii, iii, iv are ture
ii, iii, and iv are true
i and iii are true

Discuss Question

Answer: Option D. -> i and iii are true
:
D

f_{0} > f_{e}

for producing magnified image and objective aperture should be large to have good resolution.

Question 30. A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5 cm. Assuming the final image to be at the least distance of distinct vision. The magnification produced by the objective will be