Physics

Ray Optics and Mirrors

115 Questions

Ray optics covers the principles of light reflection and refraction through mirrors and lenses. The questions focus on focal length, magnification, and image formation. This physics topic is highly relevant for exams requiring science aptitude.

Concave mirror imagesConvex lens formulasFocal length calculationsMagnification ratiosImage distance

Ray Optics and Mirrors Questions

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

An object of length 2.0 cm is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. Find the size of the image if the object is at a distance of 8.0 cm from the lens.

  1. $6 cm$
  2. $4 cm$
  3. $5 cm$
  4. $1 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using 1/v - 1/u = 1/f with u = -8 cm and f = 12 cm, 1/v = 1/12 - 1/8 = (2-3)/24 = -1/24. So v = -24 cm. Magnification m = v/u = -24 / -8 = 3. Image size = m * object size = 3 * 2.0 cm = 6 cm.

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

A point object $O$ is placed on the principle axis of a convex lens of focal length $20\ cm$ at a distance of $40\ cm$ to the left of it. The diameter of the lens is $10\ cm$. If the eye is placed $60\ cm$ to the right of the lens at a distance $h$ below the principle axis, then the maximum value of $h$ to see the image will be

  1. $2.5\ cm$
  2. $5\ cm$
  3. $0\ cm$
  4. $10\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The object is at 2f (40 cm), so the image is at 2f (40 cm) on the other side. The lens diameter is 10 cm (radius 5 cm). The rays from the object pass through the lens and converge at the image point. The cone of light has a radius that scales with distance. At 60 cm from the lens (20 cm past the image), the cone radius is determined by similar triangles.

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

For two position of lens, the images are obtained on a fixed screen. If the size of the object is 2 cm and size of diminished image is 0.5 cm, the size of the other image will be

  1. 1 cm

  2. 4 cm

  3. 8 cm

  4. 16 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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


$2 \times 2=0.5 \times h$

$h=8$

option $C$ is correct

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

A lens is placed between the source of light and a wall. It forms images of area $ { A } _{ 1 }$ and  ${ A } _{ 2 }$ on the wall for its two different positions. The area of the source of light is :

  1. $ \sqrt { { A } _{ 1 }{ A } _{ 2 } } $
  2. $ \dfrac { { A } _{ 1 }+{ A } _{ 2 } }{ 2 } $
  3. $ { \left( \dfrac { \sqrt { { A } _{ 1 } } +\sqrt { { A } _{ 2 } } }{ 2 } \right) }^{ 2 } $
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


$r^{2} _{source}=r _{image1} \times r _{image2}$

$\pi r^{2} _{source}=\pi r _{image1} \times r _{image2}$

$A _{source}=\sqrt{\pi^{2} r^{2} _{image1}r^{2} _{image2}}$

$A _{source}=\sqrt{A _{1}A _{2}}$

option $A$ is correct 

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

A student focused the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under position of candle $=12.0\ cm$position of convex lens $=50.0\ cm$position of the screen $=88.0\ cm$. Where will the image be formed, if he shifts the candle towards the lens at a position of $31.0\ cm$?

  1. 19cm

  2. 48cm

  3. Infinity

  4. at center of curvature
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial positions: candle 12, lens 50, screen 88. Object distance u = 50 - 12 = 38 cm. Image distance v = 88 - 50 = 38 cm. Since u = v, 2f = 38, so f = 19 cm. If the candle is moved to 31 cm, the new object distance u = 50 - 31 = 19 cm. Since u = f, the image is formed at infinity.

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

An object is placed at a distance of $30\ cm$ from a concave lens of focal length $15\ cm$. What is the height of the object if the height of the image is $3\ cm$?

  1. $3\ cm$
  2. $1\ cm$
  3. $6\ cm$
  4. $9\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a concave lens, object distance u = -30 cm and focal length f = -15 cm. Using the lens formula 1/f = 1/v - 1/u, we find v = -10 cm. The magnification is m = v/u = -10 / -30 = 1/3. Since magnification is also height of image / height of object, and image height is 3 cm, the object height must be 3 cm divided by (1/3), which equals 9 cm.

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

In a slide show program, the image on the screen has an area 900 times that of the slide. If the distance between the slide and the screen is $x$ times the distance between the slide and the projector lens, then

  1. $x=30$
  2. $x=31$
  3. $x=500$
  4. $x=1/30$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnification of area = 900 times

So linear magnification = $\sqrt (\text{Area magnification})$ = 30 times

Let distance between slide and projector (u) be $a$

So, distance between projector and screen (v) = $m \times u = 30 a$

Distance between slide and screen = $x + 30x = 31a$

By question $ 31a = x \times  a$ 
$\implies x = 31$

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

A luminous object and a screen are at fixed distance D apart. A converging lens of focal length f is placed between the object and screen. A real image of the object in formed on the screen for two lens positions if they are separated by a distance d equal to

  1. $\sqrt {D(D+4f)}$
  2. $\sqrt {D(D-4f)}$
  3. $\sqrt {2D(D-4f)}$
  4. $\sqrt {D^2+4f}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$u+v=D$

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{D-u}+\dfrac{1}{u}=\dfrac{1}{f}$

$u^{2}-Du+Df=0$

$u _{1}= \dfrac{D+\sqrt{D(D-4f)}}{2}$ and $u _{2}=\dfrac{D-\sqrt{D(D-4f)}}{2}$

$u _{1}-u _{2}=\sqrt{D(D-4f)}$

option $B$ is correct 
Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

The distance between an object and the screen is 100 cm. A lens produces an image on the screen when the lens is placed at either of the positions 40 cm apart. The power of the lens is nearly :

  1. 3 diopter

  2. 5 diopter

  3. 2 diopter

  4. 9 diopter
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, 

$u _{1}+u _{2}=100$

$u _{1}-u _{2}=40$

=>  $u _{1}=70$ and $u _{2}=30$

for $u _{1}= -70$ $v _{1}$ will be $+30$

From lens formula, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{30}+\dfrac{1}{70}=\dfrac{1}{f}$

$\dfrac{1}{f}=\dfrac{1}{21}$

$power=\dfrac{1}{21}\times 100=5(approx)$

option $B$ is correct 
Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

A candle is placed at a distance of 20 cm from a converging lens of focal length 15 cm. The image obtained on the screen is :

  1. upright and magnified

  2. inverted and magnified

  3. inverted and diminished

  4. upright and diminished
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When an object is placed between $F$ and $2F$ then image will formed between $F$ and $2F$ on opposite side of lens  and Image formed is real, Inverted and 

magnified. 

here $F= 15 cm$ then $2F = 30 cm$

object distance $u = 20 cm$ which lies between $F$ and $2F$

therefore image formed will real, inverted and magnified.

Thus Option B is correct.

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

A light source is placed 100 cm away from a screen. A converging lens placed at a certain position between the source and the screen focuses the image of the source on the screen. The lens is moved a distance of 40 cm and it is found that it again focuses the image of the source on the screen. The focal length of the lens is :

  1. 21 cm

  2. 30 cm

  3. 40 cm

  4. 67 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The expression for focal length by displacement method is given as follows.
$f=\frac { { D }^{ 2 }-{ x }^{ 2 } }{ 4D } $
where,
D - the distance between the object and screen
x - the distance between the two positions of the lens.
Here, D = 100 cm and x = 40 cm.
So, $f=\frac { { D }^{ 2 }-{ x }^{ 2 } }{ 4D } =\frac { { 100 }^{ 2 }-{ 40 }^{ 2 } }{ 4\times 100 } =21\quad cm$.
Hence, the focal length of the lens is 21 cm.

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

A convex lens forms a real image 4 cm long on a screen. When the lens is shifted to a new position without disturbing the object or the screen, again real image is formed on the screen which is 16 cm long. The length of the object is :

  1. 8 cm

  2. 10 cm

  3. 12 cm

  4. 6cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the image sizes be $ {I} _{1} \  and \  {I} _{2} $,


By Displacement Method, object size ($OS$) is given by :
$ OS = \sqrt{{I} _{1} {I} _{2}} $

Thus, OS = $ \sqrt{64} $

$\Longrightarrow$ $OS = 8$

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

The distance between two point sources of light is 24 cm and a converging lens is kept in between two sources. The object distances of two sources from a converging lens of focal length of 9 cm, so that the image distances  of two sources are equal

  1. 12 cm

  2. 24 cm or 18cm

  3. 18 cm or 6 cm

  4. 24 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here $u _1 + u _2  =-24$.......(1).

$\dfrac{1}{v _1}-\dfrac{1}{u _1}=\dfrac{1}{9}$
and for the virtual image 
 $\dfrac{1}{-v _1}- \dfrac{1}{u _2}= \dfrac{1}{9}$

$ -(\dfrac{1}{u _1}+ \dfrac{1}{u _2})= \dfrac{2}{9} \implies u _1 u _2=108$.........(2)
On solving (1) and (2) 
We get $u^2+24u+108=0 \implies u= -18, \ - 6 $