Tag: physics

Questions Related to physics

A battery operated torch is adjusted to give a parallel beam of light. It produces illuminance of 60 lux on a wall 2m away. The illuminance produced 3 m away is 

luminous intensity measurements physics

  1. $60$ lux

  2. $\displaystyle\ \frac{80}{3}$ lux

  3. $40$ lux

  4. none of these

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Correct Option: A
Explanation:

Since $\dfrac {dF}{dA}$ is a constant, where $F$ is the illuminance and $A$ is the area of cross section. Hence the illuminance produced 3 m away does not change and it is 60 lux.

Two coherent source must have the same :

luminous intensity measurements physics

  1. amplitude

  2. phase difference

  3. frequency

  4. both B and C

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Correct Option: D
Explanation:

Two coherent source have the same :

1) Frequency
2) Phase difference.

The correct option is D.

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?

luminous intensity measurements physics

  1. $24 cm$

  2. $12 cm$

  3. $18 cm$

  4. $456 cm$

Show answer
Correct Option: A
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

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

luminous intensity measurements physics

  1. increases continuously

  2. decreases continuously

  3. first increases then decreases

  4. first decreases then increases

Show answer
Correct Option: C
Explanation:

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


Answer. C) first increases and then decreases. 

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

luminous intensity measurements physics

  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

Show answer
Correct Option: D
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.

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 

luminous intensity measurements physics

  1. $0.5$ %

  2. $1$ %

  3. $2$ %

  4. $4$ %

Show answer
Correct Option: C
Explanation:

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

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

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

$1$ % 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 

luminous intensity measurements physics

  1. $100$ lux

  2. $200$ lux

  3. $300$ lux

  4. $400$ lux

Show answer
Correct Option: B
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) $200 \ lux$

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

luminous intensity measurements physics

  1. $0.36$

  2. $0.64$

  3. $0.49$

  4. $0.51$

Show answer
Correct Option: B
Explanation:

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

Two light sources of $8 Cd$ and $12 Cd$ are placed on the same side of the photometer screen at a distance of $40 cm$ from it. Where should a $80 Cd$ source be placed to balance the illuminance?

luminous intensity measurements physics

  1. $40 cm$

  2. $60 cm$

  3. $20 cm$

  4. $80 cm$

Show answer
Correct Option: D
Explanation:

Intensity of illumination = $\dfrac {L} {d^{2}}$


Therefore
$I _1+I _2=I$

$\dfrac {8} {0.4^{2}} + \dfrac{12} {0.4^{2}} = \dfrac{80}{d^{2}}$

=>$\dfrac {20} {0.16} = \dfrac{80}{d^{2}}$

=> d = 0.8 m =80 cm

Answer. D) 80 cm

The luminous intensity of a light source is $300 Cd$. The illuminance of a surface lying at a distance of $10$ $m$ from it will be if light falls normally on it 

luminous intensity measurements physics

  1. $30$ lux

  2. $3$ lux

  3. $0.3$ lux

  4. $0.03$ lux

Show answer
Correct Option: B
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 $I = 300 cd$ and $R = 10 m$
$\implies E = \dfrac {300}{10^2} = 3   lux$