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 physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

A spherical surface of radius of curvature $R$ separates air (refractive index 1.0) from glass (refractive index 1.5).The centre of curvature is in the glass. A point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $\mathbf { O } \text { and } \mathbf { P O }= \mathrm { OQ }$.Find the distance of object from the spherical surface.

  1. 3R

  2. 5R

  3. R

  4. 2R
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

A $10\mathrm { mm }$ long awl pin is placed vertically In front of a concave mirror. A $5\ mm$ long image of the awl pin is formed at $30\mathrm { cm }$ in front of the mirror, The focal length of this mirror is 

  1. $- 30 \mathrm { cm }$
  2. $- 20 \mathrm { cm }$
  3. $- 40 \mathrm { cm }$
  4. $- 60 \mathrm { cm }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnification m = h_i/h_o = -v/u. m = -5/10 = -0.5. -0.5 = -(-30)/u => u = -60. Mirror formula: 1/f = 1/v + 1/u = 1/-30 + 1/-60 = -3/60 = -1/20. f = -20 cm.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

A point object is placed on principal axis of concave mirror of radius of curvature 10 cm at a distance 21 cm from pole of the mirror.  A glass slab of thickness 3 cm and refractive index 1.5 is placed between object and mirror $.$ Find the imaged position of the image formed. 

  1. 4

  2. 3

  3. 16.5

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

The suns diameter is $1.4\times { 10 }^{ 9 }m$ and its distance from the earth is ${ 10 }^{ 11 }m$. The diameter of its image, formed by a convex mirror of focal length 2m will

  1. 0.7 cm

  2. 1.4 cm

  3. 2.8 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a mirror, the magnification m = -v/u = f/(f-u). Given the sun is very far away (u is large), the image is formed at the focus f. The size of the image is h_i = h_o * (f/u). Plugging in values: (1.4 * 10^9) * (2 / 10^11) = 2.8 * 10^-2 m = 2.8 cm.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

The distance of real object when a concave mirror produces a real image of magnification $'m'$ is ($f$ is focal length)

  1. $\left(\frac{m - 1}{m}\right) f$
  2. $\left(\frac{m + 1}{m}\right) f$
  3. $(m-1)f$
  4. $(m+1)f$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a concave mirror, magnification m = f / (f - u). Rearranging for u: m(f - u) = f, mf - mu = f, mu = mf - f, u = f(m - 1) / m. However, for a real image, m is negative. Using m = -|m|, the distance u = f(1 + |m|) / |m|.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

Converging rays are incident on a convex spherical mirror so that their extensions intersect  $30 cm$  behind the mirror on the optical axis. The reflected rays form a diverging beam, so that their extensions intersect the optical axis  $1.2 m$  from the mirror. The focal length of the mirror is

  1. $40{ cm }$
  2. $60{ cm }$
  3. $30{ cm }$
  4. $24{ cm }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the mirror formula 1/v + 1/u = 1/f. For converging rays, u = +30 cm (virtual object). The reflected rays form a diverging beam with image at v = -120 cm. 1/(-120) + 1/30 = 1/f. 1/f = (-1 + 4) / 120 = 3/120 = 1/40. So f = 40 cm.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

The distance at which an object should be placed in front of a convex lens of focal length 10 cm to obtain a real image double the size of object will be:

  1. 30 cm

  2. 15 cm

  3. 5 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Convex lens gives the real and double-sized image when the object is placed exactly between the focus and radius of curvature.
We have, $\displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$m = \displaystyle \frac{v}{u} = 2$ or $v = 2u$


$\therefore \displaystyle \frac{1}{f} = \frac{1}{2u} - \frac{1}{-u} = \frac{1}{2u} + \frac{1}{u} = \frac{3}{2u}$

or $\displaystyle \frac{1}{10} = \frac{3}{2u}$ or $u = 15 cm$

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

A thin convex lens of focal length $30.00\ cm$ forms an image $2.00\ cm$ high, of an object at infinity. A thin concave lens of focal length $20.00\ cm$ is placed $26.00\ cm$ from the convex lens on the side of the image. The height of the image now is

  1. $1.00\ cm$
  2. $1.25\ cm$
  3. $2.00\ cm$
  4. $2.50\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The convex lens forms an image at its focal point (30 cm). The concave lens is placed 26 cm from the convex lens, so the image acts as a virtual object for the concave lens at a distance of 4 cm (30 - 26 = 4 cm). Using the lens formula 1/f = 1/v - 1/u, where f = -20 and u = +4, we get 1/v = 1/-20 + 1/4 = 4/20, so v = 5 cm. Magnification m = v/u = 5/4 = 1.25. Final height = 1.25 * 2 cm = 2.50 cm.

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

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

  1. $(I _1+I _2)/2$
  2. $I _1-I _2$
  3. $\sqrt{I _1\,I _2}$
  4. $\sqrt{I _1/I _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the displacement method, the object size O is related to the two image sizes I1 and I2 by the geometric mean formula O = sqrt(I1 * I2). This result is derived from the magnification formulas m1 = I1/O = v/u and m2 = I2/O = u/v.

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

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

  1. $I _1/I _2$
  2. $I _1\times I _2$
  3. $\sqrt{I _1\times I _2}$
  4. $\sqrt{I _1/I _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Similar to the previous question, the size of the object in the displacement method is the geometric mean of the two image sizes, O = sqrt(I1 * I2).

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.