Tag: optics

Questions Related to optics

When the distance between the object and the screen is more than 4f, we can obtain the image of the object on the screen for the two positions of the lens. It is called displacement method.In one case, the image is magnified. If $I _1$ and $I _2$ be the sizes of the two images, then the size of the object is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $(I _1+I _2)/2$

  2. $I _1-I _2$

  3. $\sqrt{I _1\,I _2}$

  4. $\sqrt{I _1/I _2}$

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Correct Option: C

If $I _1$ and $I _2$ be the size of the images respectively for the two positions of lens in the displacement method, then the size of the object is given by

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $I _1/I _2$

  2. $I _1\times I _2$

  3. $\sqrt{I _1\times I _2}$

  4. $\sqrt{I _1/I _2}$

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Correct Option: C

The distance between a point source of light and a screen which is $60$ $cm$ is increased to 180 cm. The intensity on the screen as compared with the original intensity will be : 

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $\cfrac { 1 } { 9 }$ times

  2. $\cfrac { 1 } {3 }$ time

  3. $3$ times

  4. $9$ times

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Correct Option: A

A convex lens is placed between object and a screen. The size of object is $3 cm$ and an image of height $9 cm$ is obtained on the screen. When the lens is displaced to a new position, what will be the size of image on the screen?

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $2 cm$

  2. $6 cm$

  3. $4 cm$

  4. $1 cm$

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Correct Option: D
Explanation:

The given problem is an example of displacement method, which is generally used to measure the focal length of the lens. In this method, the two image sizes and the object size are related as:

$O = \sqrt{I _1 I _2}$
$\implies 3 = \sqrt {9 \times I _2}$
$I _2 = 1\ cm$

A point object is placed on the principle axis of a converging lens and its image $(I _{1})$ is formed on its principle axis. If the lens is rotated by an small angle $\theta$ about its optical centre such that its principle axis also rotates by the same amount then the image $(I _{2})$ of the same object is formed at point $P$. Choose the correct option.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. Point $P$ lies on the new principle axis.

  2. Point $P$ lies on the old principle axis.

  3. Point $P$ is anywhere between the two principle axes

  4. None of these

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Correct Option: C

Optical axis of a thin equi-convex lens is the $X-$axis. The coordinate of a point object and its image are ($-20\ cm, 1\ cm$) and ($25\ cm,-2\ cm$) respectively:-

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. the lens is located at $x=5\ cm$

  2. the lens is located at $x=-5\ cm$

  3. the focal length of the lens is $10\ cm$

  4. the focal length of the lens is $15\ cm$

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Correct Option: B,C

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is obtained on the screen. The linear magnification of the image is 25. The lens is now moved 30 cm towards the screen and a sharp image is again formed on the screen. Find the focal length of the lens.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $1.2 cm$

  2. $14.3 cm$

  3. $14.6 cm$

  4. $14.9 cm$

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Correct Option: B

A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90 cm.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. The distance between the two positions of the lens is 30 cm.

  2. The distance of the object from the lens in its first position is 36 cm.

  3. The height of the object is 6 cm.

  4. The focal length of the lens is 21.6 cm.

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Correct Option: D
Explanation:

$h^{2} _{object}=h _{image1} \times h _{image2}$


$h _{object}=\sqrt{36}=6$

magnification of image is $\dfrac{v}{u}=\dfrac{9}{6}$

                                            $v= \dfrac{3u}{2}$

in lens displacement method , $u+v=d$ ; $uv=df$

$u+\dfrac{3u}{2}=90$

$u=36$   => $v=54$

$uv=df$ 

$f=\dfrac{36\times 54}{90}=21.6$

option $D$ is correct 

In a converging lens of focal length f and the distance between real object and its real image is 4f. If the object moves $x _1$ distance towards lens its image moves $x _2$ distance away from the lens and when object moves $y _1$ distance away from the lens its image moves $y _2$ distance towards the lens, then choose the correct option:-

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $x _1>x _2 $ and $y _1>y _2$

  2. $ x _1 < x _2 $ and $ y _1 < y _2 $

  3. $ x _1 < x _2 $ and $y _1>y _2$

  4. $x _1>x _2 $ and $y _2>y _1$

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Correct Option: A

A double convex lens is made of glass which has refractive index $1.55$ for violet rays and $1.50$ for red rays. If the focal length for violet rays is $25$ cm, the focal length for red rays will be nearly

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 37.5 cm

  2. 17.5 cm

  3. 27.5 cm

  4. 35 cm

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Correct Option: A