Physics

Ray Optics and Mirrors

141 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 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.

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

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