Optical Instruments And Some Natural Phenomena(11th And 12th > Physics ) Questions and answers for exam Preparation

The diameter of the eye-ball of a normal eye is about 2.5 cm. The power of the eye lens varies from

2 D to 10 D
40 D to 32 D
9 D to 8 D
44 D to 40 D

Explanation:-

Answer: Option D. -> 44 D to 40 D
:
D
An eye sees distant objects with full relaxation so

\frac{1}{2.5 \times 10^{- 2}} - \frac{1}{- \infty} = \frac{1}{f}

or

P = \frac{1}{f} = \frac{1}{25 \times 10^{- 2}} = 40 D

An eye sees an object at 25 cm with strain so

\frac{1}{2.5 \times 10^{- 2}} - \frac{1}{- 25 \times 10^{- 2}} = \frac{1}{f} o r P = \frac{1}{f} = 40 + 4 = 44 D

Question 2.

A man can see clearly up to 3 metres. Prescribe a lens for his spectacles so that he can see clearly up to 12 metres

$- \frac{3}{4} D$
3D
$- \frac{1}{4} D$
-4 D

Explanation:-

Answer: Option C. ->

- \frac{1}{4} D

:
C
By using

f = \frac{x y}{x - y} \Rightarrow f = \frac{3 \times 12}{3 - 12} = - 4 m

. Hence power

P = \frac{1}{f} = - \frac{1}{4} D

Question 3.

A man can see the objects up to a distance of one metre from his eyes. For correcting his eyesight so that he can see an object at infinity, he requires a lens whose power is

+0.5 D
+1.0 D
+2.0 D
–1.0 D

Explanation:-

Answer: Option D. -> –1.0 D
:
D
Hey, like you just learned, for correcting the myopic eye, the concave corrective lens should form the image for an object at infinity at the far point of the defective eye.
Applying sign convention, f = –(far point of myopic eye) = – 100 cm. So power of the lens

P = \frac{100}{f} = \frac{100}{- 100} = - 1 D

Question 4.

For a normal eye, the least distance of distinct vision is

[CPMT 1984]

0.25 m
0.50 m
25 m
Infinite

Explanation:-

Answer: Option A. -> 0.25 m
:
A

Remember this is the distance nearer to which a normal eye can't see properly, on average.

Question 5.

Image formed on the retina is

Real and inverted
Virtual and erect
Real and erect
Virtual and inverted

Explanation:-

Answer: Option A. -> Real and inverted
:
A

Image formed at retina is real and inverted.

Question 6.

A compound microscope is used to enlarge an object kept at a distance 0.03m from it’s objective which consists of several convex lenses in contact and has focal length 0.02m. If a lens of focal length 0.1m is removed from the objective, then by what distance the eye-piece of the microscope must be moved to refocus the image

2.5 cm
6 cm
15 cm
9 cm

Explanation:-

Answer: Option D. -> 9 cm
:
D
If initially the objective (focal length

F_{0}

) forms the image at distance

v_{0}

then

v_{0} = \frac{u_{0} f_{0}}{u_{0} - f_{0}} = \frac{3 \times 2}{3 - 2} = 6 c m

Now as in case of lenses in contact

\frac{1}{F_{0}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} + . . . . . = \frac{1}{f_{1}} + \frac{1}{F_{0}^{'}} {w h e r e \frac{1}{F_{0}^{'}} = \frac{1}{f_{2}} + \frac{1}{f_{3}} + . . .}

So if one of the lens is removed, the focal length of the remaining lens system

\frac{1}{F_{0}^{'}} = \frac{1}{F_{0}} - \frac{1}{f_{1}} = \frac{1}{2} - \frac{1}{10} \Rightarrow F_{0}^{'} = 2.5 c m

This lens will form the image of same object at a distance

v_{0}^{'}

such that

v_{0}^{'} = \frac{u_{0} F_{0}^{'}}{u_{0} - F_{0}^{'}} = \frac{3 \times 2.5}{(3 - 2.5)} = 15 c m

So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted i.e. 15 – 6 = 9 cm.

Question 7.

Magnifying power of a simple microscope is (when final image is formed at D = 25 cm from eye)

[MP PET 1996; BVP 2003]

$\frac{D}{f}$
$1 + \frac{D}{f}$
$1 + \frac{f}{D}$
$1 - \frac{D}{f}$

Explanation:-

Answer: Option B. ->

1 + \frac{D}{f}

:
B
This is the case when image is formed at near point.

Question 8.

The magnifying power of a simple microscope can be increased, if we use eye-piece of

Higher focal length
Smaller focal length
Higher diameter
Smaller diameter

Explanation:-

Answer: Option B. -> Smaller focal length
:
B

m=1+D/f. So, smaller the focal length, higher the magnifying power.

Question 9.

For which of the following colour, the magnifying power of a microscope will be maximum

White colour
Red colour
Violet colour
Yellow colour

Explanation:-

Answer: Option C. -> Violet colour
:
C

Question 10.

If the focal length of the objective lens and the eye lens are 4 mm and 25 mm respectively in a compound microscope. The length of the tube is 16 cm. Find its magnifying power for relaxed eye position

32.75
327.5
0.3275
None of the above

Explanation:-

Answer: Option B. -> 327.5
:
B
The image distance for the objective, in this case, is

L - f_{0} - f_{e}

as the focal lengths are not negligible compared to the length of the telescope.
By using

m_{\infty} = \frac{(L - f_{0} - f_{e}) D}{f_{0} f_{e}}

= \frac{(16 - 0.4 - 2.5) \times 25}{0.4 \times 2.5} = 327.5