Question

two monochromatic radiation x and y are incident normally on diffraction grating. The second order intensity maximum for X coincides with the third order intensity maximum for Y. What is the ration wavelength of X and wavelength Y?

Answer 1

To solve for the ratio of the wavelengths of the two monochromatic radiation \(X\) and \(Y\) when their diffraction orders coincide, we can use the diffraction grating equation:

\[
d \sin(\theta) = m \lambda
\]

where:
- \(d\) is the grating spacing (distance between adjacent slits),
- \(\theta\) is the angle of diffraction,
- \(m\) is the order of the maximum,
- \(\lambda\) is the wavelength of the radiation.

Given that the second order maximum for radiation \(X\) coincides with the third order maximum for radiation \(Y\), we can denote their wavelengths as \(\lambda_X\) for \(X\) and \(\lambda_Y\) for \(Y\).

For radiation \(X\), in the second order (\(m = 2\)):

\[
d \sin(\theta) = 2 \lambda_X
\]

For radiation \(Y\), in the third order (\(m = 3\)):

\[
d \sin(\theta) = 3 \lambda_Y
\]

Since both expressions equal the same \(d \sin(\theta)\) for the same angle \(\theta\), we can set them equal to each other:

\[
2 \lambda_X = 3 \lambda_Y
\]

Now, we can find the ratio of the wavelengths:

\[
\frac{\lambda_X}{\lambda_Y} = \frac{3}{2}
\]

Thus, the ratio of the wavelengths of \(X\) and \(Y\) is:

\[
\frac{\lambda_X}{\lambda_Y} = \frac{3}{2}
\]