Physics

Optics and Lenses

251 Questions

Enhance your physics knowledge by practicing questions on optics and the functioning of lenses. The set covers calculating combined focal lengths, lens power, and correcting vision defects like myopia and hypermetropia. These physics fundamentals are crucial for medical entrance and state board exams.

Combined focal lengthCorrecting vision defectsLens power calculationConvex and concave lensesTelescopesHuygens principle

Optics and Lenses 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

The radii of curvature of the surfaces of a double convex lens are 20 cm and 40 cm respectively, and its focal length is 20 cm. What is the refractive index of the material of the lens.? 

  1. $\dfrac{5}{2}$
  2. $\dfrac{4}{3}$
  3. $\dfrac{5}{3}$
  4. $\dfrac{4}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here $R _1$ = 20 cm, $R _2$ = -40 cm, f = 20 cm
Using lens maker's formula we get,
$\dfrac{1}{20} \, = \, (\mu \, - \, 1) \left ( \dfrac{1}{20} \, + \, \dfrac{1}{40} \right )$
$\dfrac{1}{20} \, = \, (\mu \, - \, 1) \dfrac{3}{40} \, \Rightarrow \, \mu \, = \, \dfrac{5}{3}$

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

Radius of curvature is found to be equal to twice the focal length for:

  1. Plane mirror of small aperture

  2. Spherical mirrors of small aperture

  3. Plane mirrors of large aperture

  4. Spherical mirrors of large aperture
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For spherical mirrors of small apertures, the radius of curvature is found to be equal to twice the focal length. We put this as $R = 2f$. This implies that the principal focus of a spherical mirror lies midway between the pole and centre of curvature.

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 dobleconves lens of focal length $6 cm$ is made of glass of refractive index $1.5$ the radius of curvature of of one surface is double that of other surface. The value of small radius of curvature is

  1. $6 cm$
  2. $4.5 cm$
  3. $9 cm$
  4. $4 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Lens maker's formula: 1/f = (n-1)(1/R1 - 1/R2). Given f=6, n=1.5, R1=x, R2=-2x (double-convex). 1/6 = (0.5)(1/x + 1/2x) = 0.5(3/2x) = 3/4x. So 4x = 18, x = 4.5 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

An object is placed at 15 cm from a convex lens of focal length 10 cm . Where should another convex mirror of radius 12 cm placed such that image will coincide with object

  1. 18 cm

  2. 17 cm

  3. 14 cm

  4. 20 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First, find the image position from the lens: 1/v - 1/-15 = 1/10 => 1/v = 1/10 - 1/15 = 1/30. v = 30 cm. For the mirror to make the image coincide, the light must strike the mirror normally, meaning the image must be at the center of curvature. Mirror R = 12, so C = 12. Distance = 30 - 12 = 18 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 converging bundle of light rays in the shape of cone with a vertex angle of 45 falls on a circular diaphragm of 20 cm diameter. A lens with power 5 D is fixed in the diaphragm. Diameter of face of lens is equal to that of diaphragm. If the vertex angle of new cone is 

  1. $
    \cfrac { 3 d } { 4 }
    $
  2. $
    \cfrac { 5 d } { 4 }
    $
  3. $
    2 d
    $
  4. $
    \cfrac { 3 } { 2 } d
    $
Reveal answer Fill a bubble to check yourself
B 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 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 physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

Formula of focal length in convex lens is

  1. $\displaystyle f = \frac{u+v}{u-v}$
  2. $\displaystyle f = \frac{u\times v}{u-v}$
  3. $\displaystyle f = \frac{u-v}{u+v}$
  4. $\displaystyle f = \frac{u+v}{u+v}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The standard lens formula is 1/v - 1/u = 1/f. Solving for f gives f = (uv) / (u - v).

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

A meter stick lies along the optic axis of a convex lens of focal length 40 cm its nearer end 60 cm from the mirror surface. How long is the image of stick?

  1. $200 cm$
  2. $\dfrac {200}{3} cm$
  3. $8 cm$
  4. $\dfrac{210}{6} cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the lens formula 1/f = 1/v - 1/u with f = 40 cm, the near end at u1 = -60 cm has an image distance v1 = -120 cm. The far end of the meter stick is at u2 = -160 cm, leading to an image distance v2 = -53.33 cm. The length of the image is the difference between these image positions, which evaluates to 200/3 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.