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PHYSICS, M3 EQ-Bank 9 MC

A point light source emits light uniformly in all directions. At a distance of 1 metre, a detector records a certain light intensity.

At what distance from the source would the intensity be expected to drop to \(\dfrac{1}{200}\)th of its value at 1 metre?

  1. 2 m
  2. 14 m
  3. 20 m
  4. 200 m
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\(B\)

Show Worked Solution
  • Light intensity decreases via the inverse square law: \(I_1r_1^2 = I_2r_2^2\)
  • Let \(I_1 = 200\), \(r_1 = 1\)
  • Find \(r_2\) when  \(I_2 = 1\):
\(200 \times 1^2\) \(= 1 \times r_2^2\)  
\(r_2\) \(=\sqrt{200}=14.1\ \text{m}\)  

 
\(\Rightarrow B\)

PHYSICS, M3 EQ-Bank 7 MC

A light detector measures an intensity of 180 W/m² when placed 2 metres from a light source.

What will the intensity be if the detector is moved to a distance of 6 metres from the same source?

  1. 60 Wm\(^{-2}\)
  2. 30 Wm\(^{-2}\)
  3. 20 Wm\(^{-2}\)
  4. 10 Wm\(^{-2}\)
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\(C\)

Show Worked Solution
  • Intensity of light between two points uses the inverse square law: \(I_1r_1^2 = I_2r_2^2\).
\(I_2\) \(=\dfrac{I_1r_1^2}{r_2^2}\)  
  \(=\dfrac{180 \times 2^2}{6^2}\)  
  \(=20\ \text{Wm}^{-2}\)  

 
\(\Rightarrow C\)

PHYSICS, M3 EQ-Bank 6 MC

The light intensity reaching a probe from a distant star is measured to be 10 units when the probe is 40 AU from the star.

What will the intensity be when the probe moves to a distance of 10 AU from the same star?

  1. 40 units
  2. 80 units
  3. 100 units
  4. 160 units
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\(D\)

Show Worked Solution
  • The intensity of light between two points is determined using the inverse square law: \(I_1r_1^2 = I_2r_2^2\).

\(I_2=\dfrac{I_1r_1^2}{r_2^2}=\dfrac{10 \times 40^2}{10^2}=160\ \text{units}\)

\(\Rightarrow D\)

PHYSICS, M3 EQ-Bank 3 MC

A student predicts how the intensity of light from a small lamp will decrease as the distance increases.

What percentage of light will reach a sensor placed at 5 metres from the source?

  1. \(6.25\%\)
  2. \(4\%\)
  3. \(11\%\)
  4. \(3\%\)
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\(B\)

Show Worked Solution
  • The intensity of light decreases via an inverse square law where \(I \propto \dfrac{1}{r^2}\)
  • At 5 m from the source the intensity of light measured, \(I = \dfrac{1}{5^2} = \dfrac{1}{25} = 0.04\)
  • Percentage of light reaching the sensor = 4%.

\(\Rightarrow B\)