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Physics 2 Marks Questions

Optical Instruments · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 20 questions.

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Question 12 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 22 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 32 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 42 Marks
A Galilean telescope is $27\ cm$ long when focused to form an image at infinity. If the objective has a focal length of $30\ cm,$ what is the focal length of the eyepiece?
Answer
For the given Galilean telescope$,$
$($When the image is formed at infinity$)f_0 = 30\ cm, L = 27\ cm$
Since $L = f_0 - |f_e|$
$[$Since, concave eyepiece lens is used in Galilean Telescope$]$
$\Rightarrow f_e = f_0 - L = 30 - 27 = 3\ cm$
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Question 52 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 62 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 72 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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Question 82 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 92 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 102 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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Question 112 Marks
The eyepiece of an astronomical telescope has a focal length of $10\ cm.$ 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$,F_e = 10\ cm, L = 1m = 100\ cm$
S0$, f_0 = L - f_e = 100 - 10 = 90\ cm$
and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$
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Question 122 Marks
The eyepiece of an astronomical telescope has a focal length of $10\ cm.$ 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,$F_e = 10\ cm, L = 1m = 100\ cm$
S0, $f_0 = L - f_e = 100 - 10 = 90\ cm$
and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$
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Question 132 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 142 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 152 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 162 Marks
A compound microscope forms an inverted image of an object. In which of the following cases it is kely to create difficulties?
  1. Looking at small germs.
  2. Looking at circular spots.
  3. Looking at a vertical tube containing some water.
Answer
  1. Looking at a vertical tube containing some water
Explantion:
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 172 Marks
A compound microscope forms an inverted image of an object. In which of the following cases it is kely to create difficulties?
  1. Looking at small germs.
  2. Looking at circular spots.
  3. Looking at a vertical tube containing some water.
Answer
  1. Looking at a vertical tube containing some water
Explantion:
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 182 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 192 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 202 Marks
A Galilean telescope is $27\ cm$ long when focused to form an image at infinity. If the objective has a focal length of $30\ cm,$ what is the focal length of the eyepiece?
Answer
For the given Galilean telescope, $($When the image is formed at infinity$)f_0 = 30\ cm, L = 27\ cm$
Since $L = f_0 - |f_e|$
$[$Since, concave eyepiece lens is used in Galilean Telescope$]$
$\Rightarrow f_e = f_0 - L = 30 - 27 = 3\ cm$
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2 Marks Questions - Physics STD 12 Science Questions - Vidyadip