Tag: physics

Questions Related to physics

The one 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. wavelength of the light

  3. total radiant flux entering the eye

  4. total luminous flux entering the eye

Show answer
Correct Option: D
Explanation:
Luminous flux measures the perceived power of light by adjusting the varying sensitivity to various frequencies of the human eye.

Radiant flux measures the total power of the electromagnetic radiation and does not account for eye sensitivity.

Similarly energy of light entering does not account for the fact that wavelengths beyond visible range does not contribute to brightness.

Wavelength of light alone does not give any idea about the brightness of light either.

Answer. D) total luminous 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 (nearly) by

luminous intensity measurements physics

  1. $0.5\%$

  2. $1 \%$

  3. $2\%$

  4. $4\%$

Show answer
Correct Option: C
Explanation:

Illuminance depends on inverse of area and area is proportional to $r^{2}$ where is r is distance from the source to the point .
$illuminance \propto \dfrac{1}{r^{2}}$ 
$\dfrac{\partial (illuminance)}{\partial x}= -\dfrac{2\delta r}{r}$
hence changing distance by $1$% the illuminance will differ by $-\dfrac{2\delta r\times100}{r}= -2%$%
i.e,illuminance will decrease by $2$% 
option $C$ is correct

A battery-operated torch is adjusted to send an almost parallel beam of light. It produce an illuminance of $40 \ lux$ when light falls on a wall $2 m$ away. The illuminance produced when it falls on a wall $4 m$ away is close to

luminous intensity measurements physics

  1. $40\ lux$

  2. $20\ lux$

  3. $10\ lux$

  4. $5\ lux$

Show answer
Correct Option: A
Explanation:

$Illuminance =\dfrac{ Luminous \ Power} { Area}$


Since the beam of light is parallel, so the area does not change with distance. Also the power of torch being same, the luminous power remains same.

Therefore illuminance remains same on a wall 2 m away and 4 m away.

Illuminance produced on a wall 4 m away = 40 lux

Answer. A) 40 lux

The brightness producing capacity of a source

luminous intensity measurements physics

  1. does not depend on its power

  2. does not depend on the wavelength emitted

  3. depends on its power

  4. depends on the wavelength emitted.

Show answer
Correct Option: C,D
Explanation:

The brightness producing capacity of a source "depends on its power" and "depends on the wavelength emitted" .
option $C,D$ are correct 

A lamp of luminous intensity $20$ $Cd$ is hanging at a height of $40$ $cm$ from the center of a square table of side $60$ $cm$. The illuminance at the centre of the table will be 

luminous intensity measurements physics

  1. $100$ lux

  2. $125$ lux

  3. $150$ lux

  4. none of these

Show answer
Correct Option: B
Explanation:

Luminous intensity, L = 20 Cd

Distance of source from center, d = 40 cm
Light from lamp hits the center normal ($\theta = 0 ^ {\circ}$)

Lambert's law, $I = L  Cos \theta / d^{2}$
$\longrightarrow \ I = 20  Cos 0 ^{\circ}/ (0.4)^{2}$
$\longrightarrow I = 125 lux$

Answer. B) 125 lux

Choose the correct options.

luminous intensity measurements physics

  1. Luminous flux and radiant flux have same dimensions.

  2. Luminous flux and luminous intensity have same dimensions.

  3. Radiant flux and power have same dimensions

  4. Relative luminosity is a dimensionless quantity.

Show answer
Correct Option: B,C,D
Explanation:
B.Luminous flux is luminous intensity over a solid angle. 
Thus both have dimensions of intensity.
C.Radiant flux or Radiant power is the radiant energy emitted, rreflected, transmitted or received, per unit time.
Thus both have same dimensions.
Relative luminosity is the ratio of two luminosity, it has to be dimensionless.

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

luminous intensity measurements physics

  1. $6 s$

  2. $12 s$

  3. $24 s$

  4. $48 s$

Show answer
Correct Option: C
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.

In, the visible region of the spectrum the rotation of the plane of polarization is given by $\displaystyle\theta=a+\frac{b}{\lambda^2}$. The optical rotation produced by a particular material is found to be $30^0$ per $mm$ at $\lambda=5000A^o$ and $50^0$ per $mm$ at $\lambda=4000A^o$.  the value of constant $b$ in degree $\mathring A^2$ per mm, will be

luminous intensity measurements physics

  1. $\displaystyle \frac{8}{9}\times 10^9$

  2. $\displaystyle -\frac{8}{9}\times 10^9$

  3. $\displaystyle \frac{9}{8}\times 10^8$

  4. $\displaystyle -\frac{9}{8}\times 10^8$

Show answer
Correct Option: A

An electric bulb is hanging over a table at a height of 1m above it.The illuminance on the table directly below the bulb is 40 lux. the illuminance at a point on the table 1 m away from the first point will be about

luminous intensity measurements physics

  1. 10.5 lux

  2. 14.1 lux

  3. 20.8lux

  4. 28.8 lux

Show answer
Correct Option: B
Explanation:

Let $ I $ be the intensity of bulb then

case I:$\ 40=\dfrac { I }{ 1\times 1 } \Rightarrow I=40cd$
case II: $ E=\dfrac { 40 }{ { \left( { 1 }^{ 2 }+{ 1 }^{ 2 } \right)  }^{ 3/2 } } =\dfrac { 40 }{ \sqrt { 8 }  } =14.1 lux$

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

luminous intensity measurements physics

  1. $3 s$

  2. $12 s$

  3. $24 s$

  4. $48 s$

Show answer
Correct Option: B
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).