Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Lens used to remove long-sightedness (hypermetropia) is

A. Concave lens

B. Plano-concave lens

Convexo-concave lens

Convex lens

SOLUTION

Solution : D

Question 2

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]

A. - 100 m

+ 5 m

- 5 m

Very large

SOLUTION

Solution : C

Far point of the myopic eye is the focal length. So, according to sign convention, f= -5 cm.

Question 3

A person cannot see distinctly at the distance less than one metre. Calculate the power of the lens that he should use to read a book at a distance of 25 cm

[CPMT 1977; MP PET 1985, 88; MP PMT 1990]

A. + 3.0 D

+ 0.125 D

- 3.0 D

+ 4.0 D

SOLUTION

Solution : A

$P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} .$
Applying sign convention,
$P = \frac{1}{f} = \frac{1}{- 100} - \frac{1}{- 25} = \frac{3}{100} = + 3 D .$

Question 4

For the myopic eye, the defect is cured by

[CPMT 1990; KCET (Engg.) 2000]

A. Convex lens

Concave lens

Cylindrical lens

Toric lens

SOLUTION

Solution : B

Question 5

Image is formed for the short sighted person at

[AFMC 1988]

Retina

Before retina

Behind the retina

Image is not formed at

SOLUTION

Solution : B

In short sightedness, the focal length of eye lens decreases, so image is formed before retina.

Question 6

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)

A. 200 cm

B. 160 cm

C. 2.5 cm

D. 0.1 cm

SOLUTION

Solution : 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 7

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)

A. 11.67 cm

B. 12.67 cm

C. 13.00 cm

D. 12.00 cm

SOLUTION

Solution : 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 8

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

A. 6.00 cm

B. 7.75 cm

C. 9.25 cm

D. 11.00 cm

SOLUTION

Solution : 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 9

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

A. 30 cm

B. 25 cm

C. 15 cm

D. 12 cm

SOLUTION

Solution : 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 10

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

A. +5

B. -5

C. +6

D. -6

SOLUTION

Solution : B

Magnification produced by compound microscope $m = m_{0} \times m_{e}$
where $m_{0} = ?$ and $m_{e} = (1 + \frac{D}{f_{e}}) = 1 + \frac{25}{5} = 6 \Rightarrow 30 = - m_{0} \times 6 \Rightarrow m_{0} = - 5$

Question 11

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^{\circ}

50^{\circ}

24^{\circ}

10^{\circ}

SOLUTION

Solution : C

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

Question 12

The focal lengths of the lenses of an astronomical telescope are 50 cm and 5 cm. The length of the telescope when the image is formed at the least distance of distinct vision is

A. 45 cm

B. 55 cm

\frac{275}{6} c m

\frac{325}{6} c m

SOLUTION

Solution : D

The length of the telescope, $L_{D} = f_{0} + u_{e}$ , as the objective forms the image at its focus, as the rays are from infinity, and this image acts as the object for the eyepiece.
Now, using lens formula, and using the fact that the eyepiece forms image at near point, that is $v = D (= - 25 c m)$ :
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} .$
Applying sign convention, $\frac{1}{u_{e}} = - \frac{1}{D} - \frac{1}{f_{e}} .$
Or, considering only the modulus, $u_{e} = \frac{f_{e} D}{f_{e} + D}$ .
Therefore, $L_{D} = f_{0} + u_{e} = f_{0} + \frac{f_{e} D}{f_{e} + D} = 50 + \frac{5 \times 25}{(5 + 25)} = \frac{325}{6} c m$ .

Question 13

The diameter of the moon is $3.5 \times 10^{3} k m$ and its distance from the earth is $3.8 \times 10^{5} k m$ . If it is seen through a telescope whose focal length for objective and eye lens are 4 m and 10 cm respectively, then the angle subtended by the image of the moon on the eye will be approximately

15^{\circ}

20^{\circ}

30^{\circ}

35^{\circ}

SOLUTION

Solution : B

The angle subtended by the moon on the objective of telescope $α = \frac{3.5 \times 10^{3}}{3.8 \times 10^{5}} = \frac{3.5}{3.8} \times 10^{- 2} r a d$ .
Let, the angle subtended by the image of the moon on the eye be $β$ . Also, the angle subtended by the moon on the earth, as well as the eye will be very nearly the same.
Therefore $m = \frac{f_{0}}{f_{e}} = \frac{β}{α} \Rightarrow \frac{400}{10} = \frac{β}{α} \Rightarrow β = 40 α \Rightarrow β = 40 \times \frac{3.5 \times 10^{3}}{3.8 \times 10^{5}} \times \frac{180}{π} = 20^{\circ}$ (approx).

Question 14

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

A. i, ii. iii are true

B. ii, iii, iv are ture

C. ii, iii, and iv are true

D. i and iii are true

SOLUTION

Solution : D

$f_{0} > f_{e}$ for producing magnified image and objective aperture should be large to have good resolution.

Question 15

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.

A. 100 cm

B. 10 cm

C. 20 cm

D. 25 cm

SOLUTION

Solution : A

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

Free Objective Test 02 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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