2 Marks Questions

Question 12 Marks

Can virtual image be formed on the retina in a seeing process?

Answer

The retina acts as a screen; only real images can be obtained on the screen. In case of people having eye defects, the spectacles form the virtual image of the object and the eye lens form the real and inverted image on the retina.

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Question 22 Marks

A nearsighted person cannot clearly see beyond 200cm. Find the power of the lens needed to see objects at large distances.

Answer

For the near sighted person,$\text{u}= \infty$ and $\text{v}=-200\text{cm}=-2\text{m}$
So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-2}-\frac{1}{\infty}$
$\frac{1}{\text{f}}=-\frac{1}{2}=-0.5$
So, power of the lens is -0.5D

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Question 32 Marks

A simple microscope is rated 5 for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40cm?

Answer

The simple microscope has magnification of 5 for normal relaxed eye (D = 25cm).Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$
For the relaxed farsighted eye, $\text{D} = 40\text{cm}$
So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$
So, its magnifying power is 8X.

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Question 42 Marks

A farsighted person cannot see objects placed closer to 50cm. Find the power of the lens needed to see the objects at 20cm.

Answer

For the far sighted person,u = -20cm, v = -50cm
from lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=\frac{1}{-50}-\frac{1}{-20}=\frac{1}{-20}-\frac{1}{50}=\frac{3}{100}$
$\Rightarrow\text{f}=\frac{100}{3}\text{cm}=\frac{1}{3}\text{m}$
So, power of the lens $=\frac{1}{\text{f}}=3\text{ Diopter}$

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MCQ 52 Marks

A compound microscope forms an inverted image of an object. In which of the following cases it is kely to create difficulties?

A
Looking at small germs.
B
Looking at circular spots.
✓
Looking at a vertical tube containing some water.
D
Non of this

Answer

Correct option: C.

Looking at a vertical tube containing some water.

If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.

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Question 62 Marks

A simple microscope using a single lens often shows coloured image of a white source. Why?

Answer

A simple microscope consists of a single convex lens. Sometimes due to chromatic and spherical aberrations, the image of a white source seems coloured at the corners of the lens and somewhere in between.

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Question 72 Marks

The angular magnification of a system is less than one. Does it mean that the image formed is inverted?

Answer

No, angular magnification is the ratio of the angle subtended by the final image on the eye to the angle subtended by the object on the unaided eye. Its value less than one signifies reduction in the size of the image. It does not mean that the image is inverted.

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Question 82 Marks

A Galilean telescope is 27cm long when focused to form an image at infinity. If the objective has a focal length of 30cm, what is the focal length of the eyepiece?

Answer

For the given Galilean telescope, (When the image is formed at infinity) $f_0=30 \mathrm{~cm}, \mathrm{~L}=27 \mathrm{~cm}$
Since $L=f_0-\left|f_e\right|$
[Since, concave eyepiece lens is used in Galilean Telescope]
$\Rightarrow \mathrm{f}_{\mathrm{e}}=\mathrm{f}_0-\mathrm{L}=30-27=3 \mathrm{~cm}$

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Question 92 Marks

A person wears glasses of power -2.5D. Is the person farsighted or nearsighted? What is the far point of the person without the glasses?

Answer

The person wears glasses of power -2.5DSo, the person must be near sighted
$\text{u}=\infty,$ $\text{v}=\text{far point,}$ $\text{f}=\frac{1}{-2.5}=-0.4\text{m}=-40\text{cm}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=0+\frac{1}{-40}$
$\Rightarrow \text{v}=-40\text{cm}$
So, the far point of the person is 40cm.

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Question 102 Marks

The eyepiece of an astronomical telescope has a focal length of 10cm. The telescope is focused for normal vision of distant objects when the tube length is 1.0m. Find the focal length of the objective and the magnifying power of the telescope.

Answer

For the given astronomical telescope in normal adjustment,$F_e = 10cm, L = 1m = 100cm$
S0, $f_0 = L - f_e = 100 - 10 = 90cm$
and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$

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