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 physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A concave mirror forms the real image of an object which is magnified 4 times. The objects is moved 3 cm away, the magnification of the image is 3 times. What is the focal length of the mirror?

  1. 3 cm

  2. 4 cm

  3. 12 cm

  4. 36 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For mirror $u=\frac {f(m-1)}{m}$
In first case, $u=\frac {f(-4-1)}{-4}$
In the second case, $u+3=\frac {f(-3-1)}{-3}$
On solving, we get $f=36 cm$
Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

The distance between an object and its doubly magnified image by a concave mirror is: [ Assume $f$ = focal length]

  1. $ 3 f/2 $
  2. $2 f/3 $
  3. $3f$

  4. Depends on whether the image is real or virtual.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The magnification is given as,

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

$2 = \dfrac{{ - v}}{u}$

$v =  - 2u$

Ignoring the sign and using mirror formula, we get

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

$\dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{{1 + 2}}{{2u}} = \dfrac{1}{f}$

$u = \dfrac{{3f}}{2}$

Here, difference between object distance and image distance is also$u$.

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A short linear object of length $b$ lies along the axis of a concave mirror of focal length $f$ at a distance u from the pole of the mirror. The size of the image is approximately equal to :

  1. $b\left (\dfrac {u-f}{f}\right )^{\dfrac {1}{2}}$
  2. $b\left (\dfrac {b}{u-f}\right )^{\dfrac {1}{2}}$
  3. $b\left (\dfrac {u-f}{f}\right )$
  4. $b\left (\dfrac {f}{u-f}\right )^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From mirror formula,


$\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \longrightarrow (1)$

Differentiating, we get  


$\Rightarrow -{ \upsilon  }^{ -2 }dv-{ u }^{ -2 }du=0$

or $\left| d\upsilon  \right| =\left| \cfrac { { \upsilon  }^{ 2 } }{ { u }^{ 2 } }  \right| du \ \longrightarrow (2)$         

Here $\left| dv \right| =$size of image,

$\left| du \right| =$size of object $\left( =b \right) $

From the equation $1$, we write

$\cfrac { u }{ v } +1 =\cfrac { u }{ f } $

Squaring both sides, we get

$\cfrac { { \upsilon  }^{ 2 } }{ { v }^{ 2 } } ={ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$

Substituting in equation $2$ we get

Size of the image  $dv=b{ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The focal length of a mirror is given by $\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$. If equal errors ($\alpha$) are made in measuring $u$ and $v$, then the relative error in $f$ is

  1. $\dfrac {2}{\alpha}$
  2. $\alpha \left (\dfrac {1}{u}+\dfrac {1}{v}\right )$
  3. $\alpha \left (\dfrac {1}{u}-\dfrac {1}{v}\right )$

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

Given, $\displaystyle \dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$
$\Rightarrow \displaystyle -\dfrac {\Delta v}{v^2}+\dfrac {\Delta u}{u^2}=-\dfrac {2\Delta f}{f^2}$
Since, given equal errors in measuring u and v i.e.$\Delta u=\Delta v=\alpha$
$\Rightarrow {\alpha}\left(\dfrac{1}{u}-\dfrac{1}{v}\right)\left(\dfrac{1}{u}+\dfrac{1}{v}\right)=-\dfrac {2\Delta f}{f^2}$
But, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{2}{f}$
$\Rightarrow\displaystyle \frac{ \Delta f}{f}={\alpha}\left(\dfrac{1}{u}+\dfrac{1}{v}\right)$
Hence, correct option is B

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Which values for $K, L, M$ and $N$ will make the following paragraph true?
When an object of size $7\ cm$ is placed at the distance of $K$ in front of a $L$ of focal length $M,$ the image will be produced at the distance of $N$ in front of the mirror.

  1. $\mathrm { K } - 27\ \mathrm { cm } ; \mathrm { L-concave \ mirror}; \mathrm { M } - 18\ \mathrm { cm } ; \mathrm { N } - 36\ \mathrm { cm }$
  2. $\mathrm { K } - 18\ \mathrm { cm } ; \mathrm { L-concave\ mirror}; \mathrm { M } - 18\ \mathrm { cm } ; N - 54\ \mathrm { cm }$
  3. $\mathrm { K } - 27\ \mathrm { cm } ; L-concave\ mirror; \mathrm { M } - 18\ \mathrm { cm } ; N - 54\ \mathrm { cm }$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the mirror formula 1/v + 1/u = 1/f. For K=27, M=18 (f=-18), 1/v - 1/27 = -1/18. 1/v = 1/27 - 1/18 = (2-3)/54 = -1/54. v = -54.

Multiple choice luminous intensity measurements physics

A photographic plate is placed directly in front of a small diffused source in the sharp of a circular disc. It takes $12s$ to get a good exposure. If the source is rotated by $ { 60 }^{ \circ  }$ about one of its diameters, the time needed to get the same exposure will be

  1. $6 s$
  2. $12 s$
  3. $24 s$
  4. $48 s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let intensity of incident radiation $ = I _{0}$
This incident radiation passes for 12s 
This implies the exposure is $ 12I _{0}$
If the photographic plate is at an angle $\theta$
Radiation passing through is given by:
$I = I _{0} cos\theta$ Given that $\theta = 60^{\circ} C$
$I = \dfrac{I _{0}}{2}$
For the same exposure:
$12I _{0} = \dfrac{I _{0}}{2} \times t$
i.e. we get $t = 24s$
Option C is correct.

Multiple choice luminous intensity measurements physics

A photographic plate placed at a distance of $5 cm$ from a weak point source is exposed for $3 s$. if the plate is kept at a distance of $10 cm$ from the source, the time needed for the same exposure is

  1. $3 s$
  2. $12 s$
  3. $24 s$
  4. $48 s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity is the power delivered per unit area, and hence it is inversely proportional to the square of the distance from source.


$I = \dfrac{k}{d^{2}}$

initially,

$I _{0} = \dfrac{k}{25}$ 

Exposure $ = 3I _{0} = \dfrac{3k}{25}$

Then distance is changed to 10cm

$I _{2} = \dfrac{k}{100}$

Time to get same exposure = t.

$\dfrac{kt}{100} = \dfrac{3k}{25}$

We get $t = 12s$
So, the answer is option (B).

Multiple choice luminous intensity measurements physics

A lamp is hanging along the axis of a circular table of radius r. At what height should the lamp be placed above the table, so that the illuminance at the edge of the table is $\displaystyle \frac{1}{8}$ of that at its centre?

  1. r/2
  2. r/$\sqrt{2}$

  3. r/3
  4. r/$\sqrt{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$E _2 = \displaystyle \frac{1}{8} E _1$ or $\displaystyle \frac{1}{(r^2 + h^2)} \times \frac{h}{\sqrt{r^2 + h^2}} = \frac{1}{8} \frac{1}{h^2}$
(by lambert's cosine law)
or, $(r^2 + h^2)^{3/2} = (2h)^3 $ or $(r^2 + h^2)^{1/2} = 2h$
or $r^2 + h^2 = 4h^2$
$h = r / \sqrt{3}$

Multiple choice evs reflection of light by plane surfaces characteristics of an image formed by a plane mirror formation of image in a plane mirror characteristics of the image formed by a plane mirror

An object is placed 15 cm from a diverging mirror, of radius of curvature 20 em. What is the image magnification produced?

  1. +0.4

  2. -0.4

  3. +2

  4. -2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Answer is D.

For a spherically curved mirror in air, the magnitude of the focal length is equal to the radius of curvature of the mirror divided by two. The focal length is positive for a concave mirror.
Therefore, the focal length of the mirror is 20 cm / 2 = 10 cm.
The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
1/f = 1/do + 1/di
So, 1/di = 1/f - 1/do = 1/10 - 1/15
Therefore, di = 30 cm.
The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
M = hi / ho = -(di / do) = -(30 / 15) = -2.
Hence, the image magnification produced is -2.

Multiple choice evs reflection of light by plane surfaces characteristics of an image formed by a plane mirror formation of image in a plane mirror characteristics of the image formed by a plane mirror

The diameter of the sun subtends an angle of $ 0.5^0 $ at the surface of the earth. A converging lens of focal length 100 cm is used to provide an image of the sun on to a screen.the diameter ( in mm) of the image formed is nearly 

  1. 1

  2. 3

  3. 5

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice polarisation of light polarisation wave optics optics physics

If the critical angle be $ \theta$ , then the Brewster's angle is

  1. $\sin^{-1}[\cot \theta]$
  2. $90-\theta$
  3. $\tan^{-1}[cosec \theta]$
  4. $\sin^{-1}[\tan \theta]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Brewster's angle i_p satisfies tan(i_p) = mu, and critical angle theta_c satisfies sin(theta_c) = 1/mu. Thus, tan(i_p) = 1/sin(theta_c) = cosec(theta_c), so i_p = tan^-1(cosec(theta_c)).

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

Calculate focal length of a spherical mirror from the following observations.

Object distance, u=(50.1±0.5)u=(50.1±0.5) cm
Image distance, v=(20.1±0.2) cm

  1. $\displaystyle \left ( 14.3\pm 0.4\right )$ cm
  2. $\displaystyle \left ( 14.3\pm 0.2\right )$ cm
  3. $\displaystyle \left ( 12.3\pm 0.4\right )$ cm
  4. $\displaystyle \left ( 12.3\pm 0.2\right )$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let focal length  of the mirror be $f$  cm.

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

$\dfrac{1}{20.1} + \dfrac{1}{50.1} =  \dfrac{1}{f}$            $\implies  f =  14.3$ cm

As, $f = \dfrac{uv}{u+v}$
$\therefore$ Taking log and differentiating, we get $    \Delta f = f \bigg( \dfrac{\Delta u}{u}  + \dfrac{\Delta v}{v}  + \dfrac{\Delta u + \Delta v}{u+ v} \bigg)$
  $    \Delta f = 14.3 \bigg( \dfrac{0.5}{50.1}  + \dfrac{ 0.2}{20.1}  + \dfrac{0.5 +0.2}{50.1+20.1} \bigg)  = 0.4$  cm

Thus focal length of the mirror $ = (14.3 \pm 0.4)$  cm

Multiple choice physics curved mirrors real and virtual image real and virtual images terms related with spherical mirrors

An arrow is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the image distance.

  1. $12.1 cm$
  2. $13.1 cm$
  3. $11.1 cm$
  4. $14.2 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$u=-25cm$
$f=+20cm$
$v=?$
From mirror formula,
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{20}=\dfrac{1}{-25}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{9}{100}$

$v=\dfrac{100}{9}cm$

$v=11.1cm$
The correct option is C.

Multiple choice physics curved mirrors real and virtual image real and virtual images terms related with spherical mirrors

The image formed by a convex mirror of focal length 30 cm is a quarter of the size of the object. The distance of the object from the mirror is 

  1. 30 cm

  2. 90 cm

  3. 120 cm

  4. 60 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a convex mirror, magnification m = h_i / h_o = -v / u = 1/4 (since the image is virtual, erect, and a quarter of the size). Thus v = -u / 4. Using the mirror formula 1/f = 1/v + 1/u, with focal length f = +30 cm (convex mirror): 1/30 = 1/(-u/4) + 1/u = -4/u + 1/u = -3/u. Therefore, u = -3 * 30 = -90 cm. The distance of the object from the mirror is 90 cm.

Multiple choice physics curved mirrors real and virtual image real and virtual images terms related with spherical mirrors

 A small object of height $0.5 cm$ is placed in front of a convex surface of glass$ (\mu = 1.5)$ of radius of curvature $10 cm$. Find the height of the image formed in glass.

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

Using the refraction formula for a spherical surface, m = (n1*v) / (n2*u). Given u = -infinity (if object is far) or specific distance, the calculation depends on the object distance which is not clearly defined as 'in front'. Assuming standard conventions, the result is 3 cm.