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Answer the questions.[Phy-5M]

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Question 15 Marks
i. A person suffering from myopia (near-sightedness) was advised to wear the corrective lens of power -2.5 D . A spherical lens of the same focal length was taken in the laboratory. At what distance should a student place an object from this lens so that it forms an image at a distance of 10 cm from the lens?
ii. Draw a ray diagram to show the position and nature of the image formed in the above case.
Answer
i. Given:
distance of image from the lens, $\mathrm{i}=10 \mathrm{~cm}$
power of the lens, $\mathrm{P}=-25 \mathrm{D}$
Now the focus of the lens:
$P=\frac{1}{f}$
where:
$\mathrm{f}=$ focal length
$-25=\frac{1}{f}$
$\mathrm{f}=-0.04 \mathrm{~m}=-4 \mathrm{~cm}$
From the equation of lens:
$\frac{1}{f}=\frac{1}{i}+\frac{1}{o}$
where:
o = distance of the object
$-\frac{1}{4}=\frac{1}{10}+\frac{1}{o}$
$\rho=-\frac{20}{7} \mathrm{~cm}$ i.e. negative sign means that the image formed is on the same side as that of the object.
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Question 25 Marks
An object 6 cm in size is placed at 50 cm in front of a convex lens of focal length 30 cm . At what distance from the lens should a screen be placed in order to obtain a sharp image of the object? Find the nature and size of the image. Also draw labelled ray diagram to show the image formation in this case.
Answer
Given,
$\mathrm{f}=30 \mathrm{~cm}$
$\mathrm{u}=-50 \mathrm{~cm}$
$\mathrm{h}=6 \mathrm{~cm}$
Lens formula
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v}=\frac{1}{30}-\frac{1}{50}$
$v=+75 \mathrm{~cm}$
$m=\frac{v}{u}=\frac{h_{i}}{h}$
$\frac{75}{-50}=\frac{h_{1}}{6}$
$\mathrm{h}_{\mathrm{i}}=-9 \mathrm{~cm}$
The image formed is real, increased and enlarged.
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Answer the questions.[Phy-5M] - Science STD 10 Questions - Vidyadip