Two slits in Young's experiment have widths in the ratio 1:25. The ratio of intensity at the maxima and minima in the interference pattern, I_max/I_min is:

Steps to Solve

1. Identify the Widths of the Slits Let the width of the first slit be $a_1 = 1$ (unit) and the width of the second slit be $a_2 = 25$ (units).

2. Calculate the Intensities The intensity at the maxima is given by the formula: $$ I_{max} = I_0 \left( \frac{a_1 + a_2}{2} \right)^2 $$ Where $I_0$ is the initial intensity due to a single slit.

Substituting the values: $$ I_{max} = I_0 \left( \frac{1 + 25}{2} \right)^2 = I_0 \left( \frac{26}{2} \right)^2 = I_0 \cdot 13^2 = 169 I_0 $$

3. Calculate the Intensity at Minima The intensity at the minima in a double slit experiment is given by: $$ I_{min} = I_0 \left( \frac{a_1 - a_2}{2} \right)^2 $$ Substituting the values: $$ I_{min} = I_0 \left( \frac{1 - 25}{2} \right)^2 = I_0 \left( \frac{-24}{2} \right)^2 = I_0 \cdot (-12)^2 = 144 I_0 $$

4. Find the Ratio of Intensities To find the ratio of the intensities at maxima and minima: $$ \frac{I_{max}}{I_{min}} = \frac{169 I_0}{144 I_0} = \frac{169}{144} $$