Question

The ratio of the intensities at minima to the maxima in the Young’s double slit experiment is 9:25. Find the ratio of the widths of the two slits.

Answer 1

Hint: The intensity of a light beam is the proportional to the amplitude of the wave. In Young’s double slit experiment, the minima and maxima are caused by the interference of two coherent waves at particular phase changes. The slit width determines the amplitude of the waves in the experiment.

Complete answer:
In Young’s double slit experiment, a monochromatic light source is used as the coherent waves passing through the slits. The primary wavefront splits into secondary waves at the slits. As a result of which the amplitude of the waves remains proportional to the slit width of the slits.
i.e.,
\[\text{slit width d }\propto \text{ a}\], where a is the amplitude of the secondary waves formed.
Now we know that the intensity of the bean on the screen is proportional to the square of the amplitude of the interfering beams,
i.e.,
\[\text{I}\propto {{\text{a}}^{2}}\], where I is the intensity of the beam.
For two beams interfering, the intensity is given by the square of the vector sum of the two interfering waves. i.e.,
\[\text{I}\propto {{({{\text{a}}_{1}}+{{a}_{2}})}^{2}}\].
Now, let us consider the ratio of the intensities at the maxima to the minima as given in the question –
\[\dfrac{{{I}_{\min }}}{{{I}_{\max }}}=\dfrac{9}{25}\]
 Now we know the relation between intensity and amplitude,
 \[\Rightarrow \dfrac{{{I}_{\min }}}{{{I}_{\max }}}=\dfrac{a_{1}^{2}}{a_{2}^{2}}\]
From this we can deduce the ratio of amplitude of minimum to the maximum intensities as –
\[\begin{align}
  & \Rightarrow \dfrac{a_{1}^{2}}{a_{2}^{2}}=\dfrac{9}{25} \\
 & \Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5} \\
\end{align}\] --(1)
Now we can relate the amplitude to the slit widths as –
\[a\propto {{d}_{1}}\pm {{d}_{2}}\]
Using this we can deduce that –
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{d}_{1}}-{{d}_{2}}}{{{d}_{1}}+{{d}_{2}}}\] as the ratio of minima to the maxima of amplitude.
Substituting the values of amplitude from (1),
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{3}{5}=\dfrac{{{d}_{1}}-{{d}_{2}}}{{{d}_{1}}+{{d}_{2}}}\]
Now using the componendo-dividendo rule,
\[\begin{align}
  & \dfrac{\dfrac{{{d}_{1}}}{{{d}_{2}}}-1}{\dfrac{{{d}_{1}}}{{{d}_{2}}}+1}=\dfrac{3}{5} \\
 & \Rightarrow \dfrac{{{d}_{1}}}{{{d}_{2}}}=\dfrac{5-3}{5+3}=\dfrac{2}{8}=\dfrac{1}{4} \\
\end{align}\]
i.e., the ratio of the slit width is given by 1:4.

Additional Information:
We can find the maximum and minimum amplitude of the secondary waves from the intensity of patterns using the same method.

Note:
The slit width, amplitude and intensity of the waves in an interference are all inter-related. In a Young’s double slit experiment, extreme care should be taken in measurements as even small changes in millimeters does a catastrophic change in the result.