Question

In young's double-slit experiment, the interference pattern is found to have ratio between the bright and dark fringes as $9$. This implies that

• A. the intensities at the screen due to the two slits are 5 units and 4 units,respectively
• B. the intensities at the screen due to the two slits are 4 units and 1 units respectively
• C. the amplitude ratio is 3
• D. the amplitude ratio is 2

Answer 1

Answer: (B), (D)

The correct options are
B the intensities at the screen due to the two slits are 4 units and 1 units respectively
D the amplitude ratio is 2

Let 'a' and 'b' be the amplitudes of the light from the two slits.

Thus $\frac{I_{max}}{I_{min}}=\frac{(a+b{)}^2}{(a-b{)}^2}=9$

$⟹\frac{a+b}{a-b}=3$

$⟹a=2b$

$\frac{a}{b}=2$

$\frac{I_1}{I_2}=\frac{a^2}{b^2}=4$