Physics · Science General

Optics and Light Properties

2,019 Questions

Optics and light properties focus on the behavior of light, including reflection, refraction, dispersion, and polarization. This topic also covers the wave nature of light, illumination, and the functioning of optical instruments like microscopes. Questions on these concepts are common in general science sections of SSC, Railways, and State exams.

Reflection and mirrorsRefraction and mediumsLight wave theoriesPolarization and intensity

Optics and Light Properties Questions

Multiple choice luminous intensity measurements physics

Light has a wave nature, because-

  1. the light travel in a straight line

  2. Light exhibits phenomenon of reflection and refraction

  3. Light exhibits phenomenon interference

  4. Light exhibits phenomenon of photo electric effect.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Interference is a phenomenon unique to wave motion, where waves superimpose to form a resultant wave of greater, lower, or the same amplitude. Reflection and refraction can be explained by both wave and particle models, while the photoelectric effect supports the particle nature.

Multiple choice luminous intensity measurements physics

An electric lamp and a candle produce equal illuminance on a screen when placed $80 cm$ and $20 cm$ from the screen respectively. The lamp is now covered with a thin paper which transmit $49\%$ of the luminous flux. By what distance the lamp should be moved to balance the intensities at the screen again?

  1. $24 cm$
  2. $12 cm$
  3. $18 cm$
  4. $456 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Case(i)
$\displaystyle\ \frac{I _{1}}{I _{2}}$ = $\displaystyle\ \left (\frac{80}{20} \right)^{2}$

Case(II)
$\displaystyle\ \frac{.49I _{1}}{I _{2}}$ =$\displaystyle\ \left(\frac{d}{20} \right)^{2}$

$d$ = $20\sqrt{.49(16)}$ = $20(2.8)$
 = $56$cm
Lamp be moved by $80$ - $56$ = $24$ cm

Multiple choice luminous intensity measurements physics

As the wavelength is increased from violet to red, the luminosity

  1. increases continuously

  2. decreases continuously

  3. first increases then decreases

  4. first decreases then increases
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The luminosity increases with increase in wavelenth, reaches a peak at around 550 nm then decreases.


Answer. C) first increases and then decreases. 

Multiple choice luminous intensity measurements physics

The parameter that determines the brightness of a light source sensed by an eye is 

  1. energy of light entering the eye per second

  2. wave length of the light

  3. total radiant flux entering the eye

  4. total luminous flux entering the eye
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The parameter that determines the brightness of a light source sensed by an eye is total luminous flux entering the eye.
$B = F$ where $B$ is the brightness and $F$ is the total flux entering the eye.

Multiple choice luminous intensity measurements physics

Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1% the illuminance will decrease 

  1. $0.5$ %
  2. $1$ %
  3. $2$ %
  4. $4$ %
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$E$  $\propto\displaystyle\ \frac{1}{r^{2}}$

$\displaystyle \frac{\triangle E}{E}$ $=\displaystyle\ \frac{2\triangle r}{r}$ 

$\Rightarrow$ $2$ ($1$%) $= 2$%

Multiple choice luminous intensity measurements physics

$1$ % of light of a source with luminous intensity $50 $candela is incident on a circular surface of radius $10 cm$. The average illuminance of the surface is 

  1. $100$ lux
  2. $200$ lux
  3. $300$ lux
  4. $400$ lux
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Illuminance = $\dfrac{Luminous \ intensity  \times 4\pi } {area}$

                    =$ \dfrac {50 \times 4 \pi}{100\times \pi\times0.1\times0.1} = 200 lux$

Answer B) $200 \ lux$

Multiple choice luminous intensity measurements physics

The illumination produced by A is balanced by B on the screen when B is 60 cm apart from the screen. A smoked glass plate is placed in front of A and to balance the illumination B is to move 15cm further away. Find the transmission coefficient of the smoked glass

  1. $0.36$
  2. $0.64$
  3. $0.49$
  4. $0.51$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\alpha$ = $\displaystyle\ \frac{I _{2}}{I _{2}}$ = $\displaystyle\ \left(\frac{60}{75} \right)^{2}$ = $\displaystyle\ \left( \frac{4}{5} \right)^{2}$ = $\displaystyle\ \frac{16}{25}$ = $0.64$

Multiple choice luminous intensity measurements physics

The luminous intensity of a light source is $500 Cd$. The illuminance of a surface distant $10m$ from it, will be if light falls normally on it

  1. $5$ lux
  2. $10$ lux
  3. $20$ lux
  4. $40$ lux
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit  used  when characterizing illumination conditions of a surface:
$E = \dfrac {I}{R^2}     lux$
where $I$ is the luminous intensity and $R$ is the distance.
Given $I = 500 cd$ and $R = 10 m$
$\implies E = \dfrac {500}{10^2} = 5    lux$

Multiple choice luminous intensity measurements physics

A point source of $100$candela is held $5$$m$ above a sheet of blotting paper which reflects $75$ % of light incident upon it. The illuminance of blotting paper is 

  1. $4$ phot
  2. $4$ lux
  3. $3$ phot
  4. $3$ lux
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Illuminance = $\dfrac{L}{d^{2}}$
$\dfrac{100}{5^{2}}$ = 4 lux

Answer. B) 4 lux
Multiple choice luminous intensity measurements physics

Inverse square law for illuminance is valid for 

  1. isotropic point source

  2. cylindrical source

  3. search light

  4. all types of sources
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Inverse square law for illuminance is valid and can be applied for all types of sources.

Multiple choice luminous intensity measurements physics

In the above problem, the luminance of blotting paper is 

  1. $3$ phot
  2. $3$ lux
  3. $4$ phot
  4. $4$ lux
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The source of luminance of the blotting paper is the light reflected by it. Since 75 % of light is reflected,


luminance = $\dfrac{100}{5^{2}}\times \dfrac{75}{100}$ = 3 lux

Answer. B) 3 lux

Multiple choice luminous intensity measurements physics

The light from an electric bulb is normally incident on a small surface. If the surface is tilted by $60^{0}$ from this position, then the illuminace of the surface will become

  1. half

  2. one fourth

  3. double

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

Illuminance is proportional to  Cos of angle at which light strikes the surface

If surface is tilted by $ 60^{\circ}$, then illuminace become Cos $60^{\circ}$ of original value
Cos $60^{\circ}$ = 0.5, so Illuminace becomes half
Answer. A) half

Multiple choice luminous intensity measurements physics

Light from a lamp is falling normally on a surface distant $10$ m from the lamp and the luminous intensity on it is $10$lux. In order to increase the intensity $9$ times, the surface will have to be placed at a distance of 

  1. $10$ $m$
  2. $\displaystyle\ \frac{10}{3}$ $m$
  3. $\displaystyle\ \frac{10}{9}$ $m$
  4. $10\times9$ $m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ I = \dfrac {L Cos \theta}{r^\circ}$

Since light falls normally $Cos \theta = 1$
$10 = \dfrac {L} {100}$ => L = 1000Cd

$90 = \dfrac {L} {d^{2}}$ 
 $\longrightarrow d^{2} = 1000/90$
$d= \dfrac{10}{3}$

Answer B) $\dfrac{10}{3} m$

Multiple choice luminous intensity measurements physics

An electric bulb of luminous intensity I is suspended at a height h from the center of the table having a circular surface diameter $2r$, the illuminace at the center of the circular disc will be

  1. $\displaystyle\ \frac{I}{r^{2}}$
  2. $\displaystyle\ \frac{I}{r}$
  3. $\displaystyle\ \frac{I}{h^{2}}$
  4. $\displaystyle\ \frac{I}{h}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Illuminance E at the surface, distance x away is $E= I/x^2$
here as the bulb is h distance away from the bulb. $E=I/h^2$

Multiple choice luminous intensity measurements physics

If the distance of surface from light source is doubled then the illuminance will become

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

We have illuminance,
$I \propto \dfrac {1}{R^2}$
where $R$ is the distance 
If the distance is doubled then 
$I' \propto \dfrac {I}{4}$