Physics

Light and Mirrors

458 Questions

Light and mirrors questions cover the principles of reflection, image formation, and mirror magnification. Concepts include the properties of plane mirrors, the uses of concave mirrors in solar devices, and the behavior of light rays. These fundamentals are essential general science topics in major competitive examinations.

Plane mirror propertiesConcave mirror applicationsMirror magnification formulaLight reflection lawsImage formation

Light and Mirrors Questions

Multiple choice physics superposition of waves coherence young's double slit experiment interference

In Lloyd's single mirror method we have

  1. Both sources virtual

  2. One source virtual and one real

  3. Both sources real

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By reflection we create a virtual source from real source and light from both of these sources are used to create interference pattern in Lloyd's single mirror method.

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

If David is observing his image in plane mirror. The distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror, then the distance between David and his image will be :

  1. 3 m

  2. 5 m

  3. 6 m

  4. 8 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Plane mirror produces erect image and is of same size and also at the same distance from the mirror as the object.

if he moves 1 m, distance of object from mirror= 3 m

distance of image from mirror= 3 m

distance between image and object =6 m.
Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

When the reflecting or refracting rays do not actually intersect but appear to intersect when produced backwards, 

  1. a real image is formed

  2. a virtual image is formed

  3. either a real or a virtual image is formed

  4. neither a real nor a virtual image is formed
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A virtual image is formed when the refracting or reflecting rays do not actually intersect but appear to intersect when they are produced backwards.

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

A real image is formed by

  1. actual intersection of the refracting or reflecting rays

  2. imaginary intersection of the refracting or reflecting rays produced backwards

  3. either actual or imaginary intersection of the reflecting or refracting rays

  4. neither actual nor imaginary intersection of the reflecting or refracting rays
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the refracting or reflecting rays actually intersect, a real image is formed.

Multiple choice evs light, shadows and images what makes things visible nature and sources of light light travels in straight line

Which of the following statements is true?

  1. The image formed by a pinhole camera is inverted because light travels in a straight line.

  2. Light does not change its direction on reflection.

  3. As we move the object away from the light source, the shadow of the object becomes smaller.

  4. Reflection is possible only from polished surface.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Light changes its direction after reflection. As we move the object away from the light source, the shadow becomes larger as the distance between object and the light source increases. Reflection is possible from rough surface also.

Multiple choice physics universe and space xx radio telescope and space telescopes optical instruments : telescope

The Hubble Space Telescope is designed with ____________  hyperbolic mirrors is known for good imaging performance over a wide field of view. 

  1. two

  2. four

  3. three

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

Hubble space telescope is reflecting telescope it is designed with TWO hyperbolic mirrors. A curved primary mirror is the basic optical element that creates an image at the focal plane. And secondary mirror was added to modify the optical characteristics for visual observation.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The object distance $u$ for a concave mirror:

  1. must be positive

  2. must be negative

  3. must not be negative

  4. may be negative
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Positive and negative sign depend on the assumption of sign conversion.
either side we can consider positive or negative.
Hence Option D.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The sum of the reciprocals of object distance and image distance is equal to the __________ of a mirror.

  1. focal length

  2. reciprocal of the focal length

  3. radius of curvature

  4. reciprocal of the radius of curvature
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The sum of the reciprocals of object distance and image distance is equal to the reciprocal of the focal length of a mirror.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The relation between $u,\,v\;and\;f$ for a mirror is given by

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

The relation between $u,\,v\;and\;f$ for a mirror is given by $f=\displaystyle\frac{u\times v}{u+v}$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

Choose the correct relation between $u,\,v\;and\;r$ for a spherical mirror, where $r$ is radius of curvature.

  1. $r=\displaystyle\frac{2uv}{u+v}$
  2. $r=\displaystyle\frac{2}{u+v}$
  3. $r=\displaystyle\frac{2(u+v)}{(uv)}$
  4. $r=\displaystyle\frac{2uv}{u-v}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relation between $u,\,v\;and\;f$ for a mirror is given by $f=\displaystyle\frac{u\times v}{u+v}$

so $r=\displaystyle\frac{2u\times v}{u+v}$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

State whether true or false : 

The magnification produced by a spherical mirror is the ratio of the height of the image to the height of the object.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The magnification is the ratio of height of the image to height of the object. The image is smaller than the object, the magnification will be less than 1. If the image if upside down (inverted) then magnification will be negative.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A concave mirror forms an erect image of an object placed at a distance of 10 cm from it. The size of the image is double that of the object. Where is the image formed?

  1. 20 cm behind the mirror

  2. 20 cm in front of the mirror

  3. 40 cm behind the mirror

  4. 40 cm in front of the mirror
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the image is erect , so the magnification ratio must be positive .


$m=2$

also $u=-10cm$

so $m=-\dfrac{v}{u}$

$2=-\dfrac{v}{-10}$ 

$v=20cm$

So the image is virtual and 20 cm from mirror on the other side of the object.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A dentist wants a small mirror that when placed $2$cm from a tooth, will produce $3\times$ upright image. What kind of mirror must be used and what must its focal length be?

  1. Concave mirror, $3.04$ cm
  2. Concave mirror, $1.5$ cm
  3. Convex mirror, $3.0$ cm
  4. Convex mirror, $1.5$cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$m=+\dfrac{v}{u}=\dfrac{+f}{u-f}$
$3=\dfrac{+f}{+2+f}\Rightarrow f=3cm$.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

When a ray of light parallel to the principle axis is incident on a concave mirror$,$ the reflected ray

  1. Passes through C

  2. Passes through F

  3. Passes midway between P and F

  4. retraces its path
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By definition, rays incident parallel to the principal axis of a concave mirror reflect through the principal focus (F).

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A mirror faces the negative x-axis. (Normal to its reflecting surface is$- \hat { i } )$  While a particle starts moving such that its image is formed in the mirror. At a certain instant the velocity of the particles is $3 \hat { i } + 4 \hat { j } + 5 \hat { k }$ and that of the mirror is $\hat { 1 } - \hat { j } + \hat { k }$ Choose the correct options.

  1. Magnitude of relative velocity of the image w.r.t mirrror is$\sqrt { 45 }$

  2. Magnitude of relative velocity of image w.r.t object is 4
  3. Magnitude of relative velocity of image w.r.t mirrror is $\sqrt { 45 }$
  4. Absolute velocity of the image w.r.t ground is$\sqrt { 42 }$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\begin{array}{l} In\, y-direction \ { V _{ o } }={ V _{ i } } \ \Rightarrow { V _{ i } }=4\widehat { j }  \ In\, z-direction \ { V _{ i } }={ V _{ o } } \ \Rightarrow { V _{ i } }=5\widehat { k }  \ In\, x-direction \ { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ I } }-{ V _{ M } }={ V _{ M } }-{ V _{ o } } \ \Rightarrow { V _{ I } }=2{ V _{ M } }-{ V _{ o } } \ =2\left[ { \widehat { i }  } \right] -3\left[ { \widehat { i }  } \right]  \ =-\widehat { i }  \ \therefore \overrightarrow { { V _{ I } } } =-\widehat { i } +4\widehat { j } +5\widehat { k }  \ \therefore \overrightarrow { { V _{ iM } } } =-2\widehat { i } +5\widehat { j } +4\widehat { k }  \ \therefore \left| { \overrightarrow { { V _{ iM } } }  } \right| =\sqrt { 45 }  \end{array}$

Hence,
option $(A)$ and $(C)$ are both correct answer.