Tag: physics

Questions Related to physics

In the above problem, the luminance of blotting paper is 

luminous intensity measurements physics

  1. $3$ phot

  2. $3$ lux

  3. $4$ phot

  4. $4$ lux

Show answer
Correct Option: B
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

The luminous efficiency of a lamp is $8.8$ lumen/watt and its luminous intensity is $700\ Cd$. The power of the lamp will be 

luminous intensity measurements physics

  1. $10^{1}$ $W$

  2. $10^{2}$ $W$

  3. $10^{3}$ $W$

  4. $10^{4}$ $W$

Show answer
Correct Option: C
Explanation:
$Luminous \ flux = Luminous \ intensity * solid \ angle $

Total solid angle = 4 $\pi$

So, Luminous flux = 700 x 4 $\pi$

Luminous flux = Luminous efficiency x power
=> 700 x 4 $\pi$ = 8.8 x power
=> Power of lamp = 1000 W

Answer. C) 10$^3$ W

The luminous efficiency of a lamp is $5$ lm $W^{-1}$ and its luminous intensity is $30$ candela. The power of the lamp will be 

luminous intensity measurements physics

  1. $6\pi$ $W$

  2. $12\pi$ $W$

  3. $24\pi$ $W$

  4. $48\pi$ $W$

Show answer
Correct Option: C
Explanation:

$Luminous \ flux = luminous \ intensity * 4 \pi$

$\longrightarrow$ F = $30 * 4 \pi   lm$

Luminous efficiency = $ 5  lm / W$

Power = $ F / Power$
            = $ 30 * 4 \pi / 5 = 24 \pi$

Answer. C) $24 \pi  W$

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

luminous intensity measurements physics

  1. half

  2. one fourth

  3. double

  4. four times

Show answer
Correct Option: A
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

The illuminance on screen distance $3$ m from a $100$ $W$ lamp is $25$ lm/$m^{2}$. Presuming normal incidence, the luminous intensity of the bulb will be 

luminous intensity measurements physics

  1. $100$ $Cd$

  2. $25$ $Cd$

  3. $225$ $Cd$

  4. none of these

Show answer
Correct Option: C
Explanation:

We have illuminance
$I = \dfrac {\phi}{R^2}$
$\implies \phi = I  R^2 = 25 \times 3^2 = 225 \ Cd$

A lamp of $250$ candle power is hanging at a distance of $6$m from a wall. The illuminace at a point on the wall at a minimum distance from the lamp will be 

luminous intensity measurements physics

  1. $9.64$ lux

  2. $4.69$ lux

  3. $6.94$ lux

  4. none of these

Show answer
Correct Option: D
Explanation:

1 Candle power = 0.981 Candela

Luminous intensity = 250*0.981 Candela

Minimum distance of lamp from wall = 6 m

Illuminance = $L/d^{2}$
= $250*0.981/6^{2}$
= 6.8125 lux

Answer. D) none of these

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 

luminous intensity measurements physics

  1. $10$ $m$

  2. $\displaystyle\ \frac{10}{3}$ $m$

  3. $\displaystyle\ \frac{10}{9}$ $m$

  4. $10\times9$ $m$

Show answer
Correct Option: B
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) $\dfrac{10}{3} m$

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

luminous intensity measurements physics

  1. $\displaystyle\ \frac{I}{r^{2}}$

  2. $\displaystyle\ \frac{I}{r}$

  3. $\displaystyle\ \frac{I}{h^{2}}$

  4. $\displaystyle\ \frac{I}{h}$

Show answer
Correct Option: C
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. $E=I/h^2$

The illuminance of a surface distance $10$ m from a light source is $10$ lux. The luminous intensity of the source for normal incidence will be 

luminous intensity measurements physics

  1. $10^{1} Cd$

  2. $10^{2} Cd$

  3. $10^{3} Cd$

  4. none of these

Show answer
Correct Option: C
Explanation:

We have illuminance
$I = \dfrac {\phi}{R^2}$
$\implies \phi = I  R^2 = 10 \times 10^2 = 10^3  Cd$

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

luminous intensity measurements physics

  1. $\displaystyle\ \frac{1}{2}$ times

  2. $2$ times

  3. $\displaystyle\ \frac{1}{4}$ times

  4. $4$ times

Show answer
Correct Option: C
Explanation:

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