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

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 : 

  1. $\cfrac { 1 } { 9 }$ times
  2. $\cfrac { 1 } {3 }$ time
  3. $3$ times
  4. $9$ times
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Intensity is inversely proportional to the square of the distance from a point source (I proportional to 1/r^2). The distance increases from 60 cm to 180 cm, a factor of 3. Thus, the intensity changes by a factor of 1/(3^2) = 1/9.

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

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?

  1. $2 cm$
  2. $6 cm$
  3. $4 cm$
  4. $1 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
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:

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

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

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.

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a lens is rotated about its optical center, the image of a point object on the principal axis moves in a circular arc centered at the optical center. The new image position P will not lie on either the original or the new principal axis.

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

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.

  1. $1.2 cm$
  2. $14.3 cm$
  3. $14.6 cm$
  4. $14.9 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the displacement method formulas: m1 = 25, m2 = 1/25 (since the lens is moved). The distance between positions is d = 30 cm. The formula for focal length is f = (D^2 - d^2) / 4D. Alternatively, using magnification m = (D-d)/2f, one can solve for f.

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

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.

  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.
Reveal answer Fill a bubble to check yourself
D Correct answer
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 

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

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:-

  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 $y _2>y _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a lens with object-image distance 4f, the magnification m = -1 at the center. Near this point, the displacement of the image is greater than the displacement of the object (m > 1 or m < -1), but the question asks about relative movements. Specifically, for a real object and real image, the longitudinal magnification is m^2. Since m^2 > 1 for positions away from 2f, the image moves more than the object.

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

The diameter of the sun is $1.4 \times 10 ^ { 9 } \mathrm { m }$ and its distance from the earth is $1.5 \times 10 ^ { 11 } \mathrm { m } .$ The radius of the image of the sun formed by a lens of focal length $20 \mathrm { cm }$ is

  1. $93 mm$
  2. $0.093mm$
  3. $9.3 mm$
  4. $0.93 mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The image size h_i = f * tan(theta), where theta is the angular diameter of the sun. tan(theta) = diameter / distance = 1.4e9 / 1.5e11 = 1.4/150. h_i = 20 cm * (1.4/150) = 200 mm * 0.00933 = 1.86 mm. The radius is half of this, 0.93 mm.