Free Objective Test 01 Practice Test - 11th and 12th

Question 1

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

A. 2 D to 10 D

B. 40 D to 32 D

C. 9 D to 8 D

D. 44 D to 40 D

SOLUTION

Solution : 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

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

[CPMT 1984]

0.25 m

0.50 m

25 m

Infinite

SOLUTION

Solution : A

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

Question 3

Image formed on the retina is

A. Real and inverted

Virtual and erect

Real and erect

Virtual and inverted

SOLUTION

Solution : A

Image formed at retina is real and inverted.

Question 4

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

A. +0.5 D

B. +1.0 D

C. +2.0 D

D. –1.0 D

SOLUTION

Solution : 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 5

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

B. 3D

- \frac{1}{4} D

D. -4 D

SOLUTION

Solution : 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 6

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

A. Higher focal length

Smaller focal length

Higher diameter

Smaller diameter

SOLUTION

Solution : B

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

Question 7

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

White colour

Red colour

Violet colour

Yellow colour

SOLUTION

Solution : C

Question 8

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}

SOLUTION

Solution : B

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

Question 9

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

A. 2.5 cm

B. 6 cm

C. 15 cm

D. 9 cm

SOLUTION

Solution : 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 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

A. 32.75

B. 327.5

C. 0.3275

D. None of the above

SOLUTION

Solution : 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 $m_{\infty} = \frac{(L - f_{0} - f_{e}) D}{f_{0} f_{e}}$
$= \frac{(16 - 0.4 - 2.5) \times 25}{0.4 \times 2.5} = 327.5$

Question 11

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 [MP PMT 1991]

200 cm

160 cm

2.5 cm

0.1 cm

SOLUTION

Solution : C

Question 12

In a compound microscope, if the objective produces an image $I_{o}$ and the eye piece produces an image $I_{e}$ , then

I_{o}

is virtual but

I_{e}

is real

$I_{o}$ is real but $I_{e}$ is virtual

$I_{o}$ and $I_{e}$ are both real

$I_{o}$ and $I_{e}$ are both virtual

SOLUTION

Solution : B

In a compound microscope, objective forms real image while an eye piece forms a virtual image.

Question 13

When the length of a microscope tube increases, its magnifying power [MNR 1986]

A. Decreases

Increases

Does not change

D. May decrease or increase

SOLUTION

Solution : A

In compound microscope objective forms real image while eye piece forms virtual image.

Question 14

The focal length of objective and eye lens of a microscope are 4 cm and 8 cm respectively. If the least distance of distinct vision is 24 cm and object distance is 4.5 cm from the objective lens, then the magnifying power of the microscope will be

B. 32

C. 64

D. 20

SOLUTION

Solution : B

Question 15

Focal length of the objective and eyelenses of a telescope are 200 cm and 5 cm. The maximum magnifying power of the telescope is

A. 40

B. 48

C. 60

D. 100

SOLUTION

Solution : B

$m = \frac{f_{0}}{f_{e}} [1 + \frac{f_{e}}{D}] m = \frac{200}{5} [1 + \frac{5}{25}] = 40 [1 + \frac{1}{5}] = 40 \times \frac{6}{5} = 8 \times 6 m = 48$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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