Question

The maximum number of possible interference maxima for slit separation equal to 1.8 $\lambda$, where $\lambda$ is the wavelength of light used, in a Young’s double slit experiment is

• A. zero
• B. 3
• C. infinite
• D. 5

Answer 1

Answer: (B)

As $sin\,\theta =\frac{n\lambda}{d}$ and $sin \theta$ cannot be $≯ 1$
$\therefore \, 1 = \frac{n\lambda}{1.8\lambda}$
or $n= 1.8$
Hence maximum number of possible
interference maximas, $0, \pm 1\, i.e.\, 3$