Tag: luminous intensity

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Questions Related to luminous intensity

Multiple choice luminous intensity measurements physics

Choose the correct options.

  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.
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
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.
Multiple choice luminous intensity measurements physics

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

  1. $6 s$
  2. $12 s$
  3. $24 s$
  4. $48 s$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice luminous intensity measurements physics

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

  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. $\displaystyle -\frac{9}{8}\times 10^8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given theta = a + b/lambda^2. For lambda1 = 5000 A, theta1 = 30. For lambda2 = 4000 A, theta2 = 50. Two equations: 30 = a + b/(5000^2) and 50 = a + b/(4000^2). Subtracting: 20 = b * (1/16*10^6 - 1/25*10^6) = b * (9 / 400*10^6). b = 20 * 400*10^6 / 9 = 8000/9 * 10^6 = 8/9 * 10^9.

Multiple choice luminous intensity measurements physics

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

  1. 10.5 lux

  2. 14.1 lux

  3. 20.8lux

  4. 28.8 lux
Reveal answer Fill a bubble to check yourself
B Correct answer
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$

Multiple choice luminous intensity measurements physics

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

  1. $3 s$
  2. $12 s$
  3. $24 s$
  4. $48 s$
Reveal answer Fill a bubble to check yourself
B Correct answer
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).

Multiple choice luminous intensity measurements physics

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

  1. continuously increases

  2. continuously decreases

  3. increases, then decreases

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

With increase in wavelength of visible light, the sensitivity to eye increases, reaches a peak at 555 nm and then decreases. So luminosity increases then decreases.


Answer. C) Increases, then decreases

Multiple choice luminous intensity measurements physics

The brightness of a source based upon sensation of eye is determine by:

  1. radiant flux entering the eye

  2. luminous flux entering the eye

  3. wavelength of light

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Only light of visible range excites the retina of eyes. The range Light, which can human eyes sense is called Luminus of light.
Hence, (b) is correct.

Multiple choice luminous intensity measurements physics

The lumen efficiency, if an electric bulb emit $68.5\dfrac{lumen}{watt}$ is:

  1. $2.5\%$
  2. $5\%$
  3. $10\%$
  4. $20\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Luminous efficiency is the ratio of luminous flux to power. A standard 100% efficient source (monochromatic green light at 555 nm) produces 685 lumens/watt. Efficiency = (68.5 / 685) * 100% = 10%.

Multiple choice luminous intensity measurements physics

A surface is receiving light normally from a source, which is at a distance of $8\ m$ from it. If the source is moved closer towards the surface, so that the distance between them becomes $4\ m$, then the angle through which the surface may be turned so that illuminance remain as it was: 

  1. $\theta=\cos^{-1}(1/3)$
  2. $\theta=\cos^{-1}(1/4)$
  3. $\theta=\cos^{-1}(1/8)$
  4. $\theta=\cos^{-1}(1/6)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Illuminance E = (I / d^2) * cos(theta). Initially, E = I / 8^2 = I / 64. Finally, E = (I / 4^2) * cos(theta) = (I / 16) * cos(theta). Equating them: I / 64 = I / 16 * cos(theta) => cos(theta) = 16 / 64 = 1/4. Thus, theta = cos^-1(1/4).

Multiple choice luminous intensity measurements physics

A point source generates $10\ J$ of light energy in $2\ s$. The luminous flux of source is:

  1. $5$ lumen
  2. $10$ lumen
  3. $50$ lumen

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

Luminous flux $=\dfrac{light\ energy}{time}=\dfrac{10}{2}=5$ lumen