Inverse Square Law Questions

It states that “the intensity of the light to an observer from a source is inversely proportional to the square of the distance from the observer to the source”.

Inverse square law is given by the formula:

\(\begin{array}{l}I\alpha\frac{1}{d^{2}}\end{array} \)

Read more: Inverse-square law.

Important Inverse Square Law Questions with Answers

1. State true or false: Inverse-square law aids in calculating various astronomical distances.

  1. TRUE
  2. FALSE

Answer: a) TRUE

Explanation: Inverse-square law helps to calculate astronomical distances.

2. What is the SI unit of the intensity of light?

  1. Watt
  2. Joules
  3. Lumen
  4. Radians

Answer: c) Lumen

Explanation: The SI unit of the intensity of light is the lumen.

3. Bending of the light is referred to as ______

  1. Reflection
  2. Refraction
  3. Interference
  4. Polarization

Answer: b) Refraction

Explanation: The phenomenon of bending of light is termed as refraction.

4. When the brightness of light when moved away from the source ______

  1. The brightness increases
  2. The brightness decreases
  3. The brightness remains constant
  4. None of the options

Answer: b) The brightness decreases

Explanation: The brightness decreases when the light is moved away from the source.

5. The formula of inverse square law is _____

  1. \(\begin{array}{l}I\alpha{d^{2}}\end{array} \)
  2. \(\begin{array}{l}I\alpha\frac{d^{2}}{0}\end{array} \)
  3. \(\begin{array}{l}I\alpha\frac{1}{d^{2}}\end{array} \)
  4. \(\begin{array}{l}I\alpha{1}+{d^{2}}\end{array} \)

Answer: c)

\(\begin{array}{l}I\alpha\frac{1}{d^{2}}\end{array} \)

Explanation:

\(\begin{array}{l}I\alpha\frac{1}{d^{2}} \textup{ is the formula to calculate the inverse square law.}\end{array} \)

6. Consider intensity of lights I1 and I2 at the distances d1 and d2.

  1. \(\begin{array}{l}\frac{I_{1}}{I_{2}}\alpha \frac{d_{2}^{2}}{d_{1}^{2}}\end{array} \)
  2. \(\begin{array}{l}\frac{I_{2}}{I_{1}}\alpha \frac{d_{2}^{2}}{d_{1}^{2}}\end{array} \)
  3. \(\begin{array}{l}\frac{I_{2}}{I_{1}}\alpha \frac{d_{2}^{2}}{d_{2}^{1}}\end{array} \)
  4. \(\begin{array}{l}\frac{I_{2}}{I_{1}}+ \frac{d_{2}^{2}}{d_{1}^{2}} = 0\end{array} \)

Answer:

\(\begin{array}{l}\frac{I_{1}}{I_{2}}\alpha \frac{d_{2}^{2}}{d_{1}^{2}}\end{array} \)

Explanation: the inverse square law for light sources of intensity I1 and I2 at the distances d1 and d2 is given by the formula,

\(\begin{array}{l}\frac{I_{1}}{I_{2}}\alpha \frac{d_{2}^{2}}{d_{1}^{2}}\end{array} \)

7. How is the inverse square law used in the X-ray technique?

  1. To calculate the thickness of the film
  2. To find distance between the source to film
  3. To calculate the intensity of the x-rays
  4. To calculate the frequency of light

Answer: b) To find distance between the source to film

Explanation: The inverse square law helps in finding the source to film distances.

8. What is a refractometer?

It is a device that measures the concentration of a particular substance in a liquid solution.

9. State true or false: For the light to undergo the process of interference, coherent sources of light are needed.

  1. TRUE
  2. FALSE

Answer: a) TRUE

Explanation: Amplitudes and intensities must be nearly equal and coherent sources are required for the light to undergo interference.

10. The light waves which travel in a single plane are known as _____

  1. Un-polarized light waves
  2. Polarized light waves
  3. Non-polarized light waves
  4. None of the options

Answer: b) Polarized light waves

Explanation: Polarized light waves travel in a single plane.

Practice Questions

  1. What is light?
  2. What are the various phenomena that light can undergo?
  3. State the inverse square law.
  4. What is the formula of inverse-square law?
  5. List the applications of the inverse square law.

Related links

Visible Light

Inverse Square Law Formula

De-Broglie’s equation

Inverse Square Law Calculator

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