Problem 19 - Master Resource Book in JEE Main Physics

Question

A parallel beam of light of intensity \(I_{0}\) is incident on a glass plate, \(25 \%\) of light is reflected by upper surface and \(50 \%\) of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is (a) \(\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^{2}\) (b) \(\left(\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^{2}\) (c) \(\frac{5}{8}\) (d) \(\frac{8}{5}\)

Step-by-Step Solution

Answer

The correct answer is (a).

Step 1: Compute Light Intensities

Calculate the intensity of light reflected by the upper surface, which is \(25\%\) of \(I_{0}\), i.e., \(I_{1} = 0.25 I_{0}\). Calculate the intensity of light incident on the lower surface. This is the remaining \(75\%\) of \(I_{0}\), i.e., \(0.75 I_{0}\). The intensity of light reflected from the lower surface is \(50\%\) of the light incident on it, hence \(I_{2} = 0.50 \times 0.75 I_{0} = 0.375 I_{0}\).

Step 2: Determine Interference Terms

Interference involves combining the two reflected intensities. If \(I_{1}\) and \(I_{2}\) interfere, they can be combined based on their amplitudes. From intensity, amplitude \(a\) is proportional to the square root of intensity \(I\). Calculate \(a_{1} = \sqrt{0.25 I_{0}}\) and \(a_{2} = \sqrt{0.375 I_{0}}\).

Step 3: Calculate Superposition Intensity Conditions

For constructive interference (maximum intensity), \(I_{\text{max}} = (a_{1} + a_{2})^2\). For destructive interference (minimum intensity), \(I_{\text{min}} = (a_{1} - a_{2})^2\).

Step 4: Calculate Maximum Intensity

Compute \(I_{\text{max}} = \left(\sqrt{0.25 I_{0}} + \sqrt{0.375 I_{0}}\right)^2 = I_{0} \left(\frac{1}{2} + \sqrt{\frac{3}{8}} \right)^2\).

Step 5: Calculate Minimum Intensity

Compute \(I_{\text{min}} = \left(\sqrt{0.25 I_{0}} - \sqrt{0.375 I_{0}}\right)^2 = I_{0} \left(\frac{1}{2} - \sqrt{\frac{3}{8}} \right)^2\).

Step 6: Compute Ratio of Intensities

The ratio of maximum to minimum intensity is \(\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2} - \sqrt{\frac{3}{8}}}\right)^2\).

Key Concepts

Intensity Calculations

Intensity calculations help us understand how much light is present in a given area. In optics, the term often refers to the amount of light energy hitting a surface. In the given problem, we have a beam of light with an initial intensity of \( I_{0} \). This beam hits a glass plate, which reflects parts of the light from its surfaces.
The first calculation involves the light reflected off the upper surface of the glass. Here, \( 25\% \) of the light, or \( 0.25 I_{0} \), is reflected. The remaining light, \( 0.75 I_{0} \), makes it to the lower surface. From this, \( 50\% \) is reflected back, resulting in an intensity of \( I_{2} = 0.375 I_{0} \). Calculating these intensities is crucial, as these values are used to assess the interference pattern that these reflections create.
Intensity calculations always consider the fraction of light energy that gets reflected or transmitted and directly influence how bright or dim the resultant light would be seen.

Constructive Interference

Constructive interference occurs when two waves meet in such a way that their crests and troughs align, reinforcing each other. This results in a wave with greater amplitude, meaning more intensity or brightness in the case of light waves. In the context of this problem, the two reflected light intensities \( I_{1} \) and \( I_{2} \) combine to form this effect.
To understand the maximum intensity resulting from constructive interference, we focus on combining the amplitudes of the two interfering waves. Since the intensity is proportional to the square of its amplitude, the amplitude \( a \) must be square-rooted. Calculating the intensity for constructive interference involves:

• Finding individual amplitudes: \( a_{1} = \sqrt{0.25 I_{0}} \) and \( a_{2} = \sqrt{0.375 I_{0}} \).
• Summing these amplitudes: \((a_{1} + a_{2})^2\).

Thus, the maximum intensity (\( I_{\text{max}} \)) is derived as \( \left(\frac{1}{2} + \sqrt{\frac{3}{8}} \right)^{2} \times I_{0} \). Constructive interference is a fascinating phenomenon where light waves "fuse" to form a more luminous result.

Destructive Interference

Destructive interference is the process where two waves meet out of sync—meaning that the crest of one wave meets the trough of another. This causes them to cancel each other out partially or wholly, leading to reduced amplitude and, thus, less intensity. In this exercise, this concept shows the minimum intensity when combining the two reflected light intensities.
To compute the minimum intensity due to destructive interference:

• As before, first compute the amplitudes: \( a_{1} = \sqrt{0.25 I_{0}} \) and \( a_{2} = \sqrt{0.375 I_{0}} \).
• Then, subtract the amplitudes: \((a_{1} - a_{2})^2\).

This operation gives us the minimum intensity \( I_{\text{min}} = \left(\frac{1}{2} - \sqrt{\frac{3}{8}} \right)^2 \times I_{0} \). Destructive interference explains why sometimes, two light waves can meet and result in darkness, demonstrating how waves can effectively reduce each other's impact.