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 in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A concave lens of focal length $f$ produces an image $(1/x)$ of the size of the object. The distance of the object from the lens is

  1. $(x - 1)f$
  2. $(x + 1)f$
  3. $\dfrac{(x - 1)}{x}$ $f$
  4. $\dfrac{(x + 1)}{x}$ $f$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a concave lens, magnification m = h_i / h_o = 1/x = f / (u + f). Solving for object distance u yields u = (x - 1)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 thin. rod of length f/ 3 is placed along the principal axis of a concave mirror of focal length f such that its image which is real and elongated, just touches one end of the rod. What is its magnification ?

  1. +2

  2. -3

  3. -1.5

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

The rod of length f/3 is placed along the axis. Let the ends be at u1 and u2. The image ends are at v1 and v2. Magnification m = dv/du. Using 1/v + 1/u = 1/f, differentiating gives dv/v^2 = -du/u^2, so m = -v^2/u^2. Given the geometry, the calculation leads to m = -1.5.

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

Let the equation connecting object distance $u$, image distance $v$ and focal length $f$ for a lens be $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$. A student measures values of $u$ and $v$, with their associated uncertainties.
These are $u = 50\ mm \pm 3\ mm, v = 200\ mm \pm 5\ mm$. He calculates the value of $f$ as $40\ mm$. What is the uncertainty in this value?

  1. $\pm 2.1\ mm$
  2. $\pm 3.4\ mm$
  3. $\pm 4.5\ mm$
  4. $\pm 6.8\ mm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 1/f = 1/u + 1/v, the uncertainty is df/f^2 = du/u^2 + dv/v^2. Plugging in the values: df = f^2 * (du/u^2 + dv/v^2) = 40^2 * (3/50^2 + 5/200^2) = 1600 * (3/2500 + 5/40000) = 1600 * (0.0012 + 0.000125) = 1600 * 0.001325 = 2.12 mm.

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 man has a concave shaving mirror of focal length $0.2$ m. How far should the mirror be held from his face in order to give an image of two fold magnification? 

  1. $0.1$ m
  2. $0.2$ m
  3. $0.3$ m
  4. $0.4$ m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Concave shaving mirror  $f = 0.2m$

Magnification,$m = 2$
$m = \dfrac{-v}{u}$

$2 = \dfrac{-v}{u}$
$v = -2u$
Using mirror formula 
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$- \dfrac{1}{0.2} = -\dfrac{1}{2u} + \dfrac{1}{u}$

$-\dfrac{1}{0.2}=\dfrac{1}{2u}$

$u = -0.1m$
i.e shaving mirror should be 10cm ahead of man

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

For position of real object at $x _1$ and $x _2 (x _2 > x _1)$ magnification is equal to $2$. Find out $\dfrac{x _1}{x _2}$. if focal length of converging lens $f = 20 \,cm$.

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

$m = \left(\dfrac{f}{f + u}\right)$

$-2 = \dfrac{20}{20 - x _2}$

$-10 x _2 = 10$
$x _2 = 20 \,cm$

$m = 2 = \dfrac{20}{20 - x _1}$

$20 - x _1 = 10$
$x _1 = 10$

$\dfrac{x _1}{x _2} = \dfrac{10}{20} = \left(\dfrac{1}{2}\right)$

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 convex lens of focal length 0.12 m produces a virtual n image which is thrice the size of the object. Find the distance between the object and the lens

  1. 0.04 m

  2. 0.08 m

  3. 0.12 m

  4. 0.24 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From lens formula
$\dfrac{{1}{v}-\dfrac{1}u}=\dfrac{1}{f}$
$\dfrac{u}{v}-1=\dfrac{u}{f}$
$\dfrac{u}{v}=\dfrac{u+f}{f}$
$\dfrac{v}{u}=\dfrac{f}{u+f}$
For a convex lens f = 0.12m and m=3 then find u
So ,  $3=\dfrac{0.12}{0.12+u}$
$0.36+3u = 0.12$
$u=-0.08 m$

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 $2.0 cm$ object is placed $15 cm$ in front of a concave mirror of focal length $10 cm$. What is the size and nature of the image?

  1. $4 cm$, real
  2. $4 cm$, virtual
  3. $1.0 cm$, real

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

$\dfrac { 1 }{ v } -\dfrac { 1 }{ 15 } =\dfrac { 1 }{ -10 } $
$\Rightarrow v=-30cm$
$\therefore m=-\dfrac { v }{ u } =-\dfrac { -30 }{ 15 } =2$
So, image size will be $4 cm$ (real)

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 thin rod of length $\dfrac {f}{3}$ is placed along the optic axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. The magnification is:

  1. $\dfrac {3}{4}$
  2. $\dfrac {1}{2}$
  3. $\dfrac {3}{2}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, the object lies along the axis. The two ends of the object should be treated as two point objects and the difference between the corresponding image distances gives the length of the image. When one end of the image touches the rod, this end must be at 2 f. In this situation the other end of the rod can be towards the left or right of 2 f. Since the image of the rod is elongated, the other end of the rod must lie between f and 2 f, the image (when the object lies between f and 2f, the image is formed more far away behind 2 f).
So, object distance for closer end of the rod is 2f-f/3 and that of the farther end is 2 f. The difference between the corresponding image distance is found to be $\dfrac {f}{2},$ i.e. length of the image is $\dfrac {f}{2}$.
magnification$=\dfrac {\text {length of image}}{\text {length of object}}=\dfrac {f/2}{f/3}=\dfrac {3}{2}$
Note that the image is elongated and the only option which is greater than 1 is (c).

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

An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is then

  1. 1 m

  2. 0.5 m

  3. 0.24 m

  4. 2 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $m =\dfrac{v}{u} =-4$

$\Rightarrow u = \dfrac{-v}{4}$------------(1)

Also, $|u| + |v| = 1.5$

$\dfrac{v}{4} + v = 1.5$

$\Rightarrow \dfrac{(v+4 v)}{4} = 1.5$

 $\Rightarrow  v = 1.2 m$

So, putting the value of $v$ in equation (1):
 $u= \dfrac{-1.2}{4} = -0.3 m\,\,\,$

$\therefore  f =\dfrac{uv}{u-v}$

$\Rightarrow f=\dfrac{(-0.3 \times 1.2)}{(-0.3 - 1.2)} = 0.24m$
Hence the correct option is $(C)$

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 expression for the magnification of a spherical mirror in the terms of focal length (f) and the distance of the object from mirror (u) is

  1. $\frac{-f}{u-f}$
  2. $\frac{f}{u+f}$
  3. $\frac{-f}{u+f}$
  4. $\frac{f}{u-f}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Equation of spherical mirror is $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ , where v is the image distance.
solving,
Replacing $v$ with $v=mu$ , 
$\dfrac{1}{f} = \dfrac{1}{u}  (1 + \dfrac{1}{m} ) $
 $u = f (\dfrac{1}{m} +1)$ where m = magnification $= \dfrac{v}{u}$
$u =\dfrac{ f}{m} + f$
$m = \dfrac{f }{ (u - f)}$
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 $L$ lies on the axis of a spherical mirror of focal length $f$ at a distance $u$ from the mirror. Its image has an axial length $L$ equal to :

  1. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 1/2 }$
  2. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 1/2 }$
  3. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 2 }$
  4. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From mirror formula,       $\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } $


On differentiating, we get


      $\cfrac { -dv }{ { v }^{ 2 } } -\cfrac { du }{ { u }^{ 2 } } =0\\ \therefore dv=-du{ \left( \cfrac { v }{ u }  \right)  }^{ 2 }\\ as\quad \cfrac { v }{ u } =\cfrac { f }{ u-f } \\ \therefore dv=-du{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }\\ { L }^{ \prime  }=L{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }$

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

If an object is placed at a distance of 20cm from the pole of a concave mirror, the magnification of its real image is 3. If the object is moved away from the mirror by 10cm, then the magnification is -1.

  1. True

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

$M= \frac{f}{f-d _0}$ and real image has M negative

$-3= \frac{f}{f-20}$

$-3f+60=f$

$f=15 cm$

$M= \frac{15}{15-30}$

$M= -1$

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 convex lens is given, for which the minimum distance between an object and its rel image is $40cm$. An object is placed at a distance of $15cm$ from this lens. The liner magnification of adjustment will be 

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

Given object distance $u=15$ cm

Distance between object and real image produced $=40 $cm
Thus image distance $v=40-15=25$ cm
Also we know linear magnification,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$