Question

The maximum number of possible interference maxima for slit-separation equal to four times the wavelength in Young's double-slit experiment is

Answer 1

Step 1: Formula used

We know, for Young's double-slit experiment, $d\sin \theta =n\lambda$

$n\to$Order of maxima

$\lambda \to$Wavelength of light

$d\to$ The separation between two slits

$\theta \to$The angle between the path and a line from the slits to the screen

Step 2: Substitute the given condition in the equation

Given, slit-separation is equal to four times the wavelength, $d=4\lambda$

$\Rightarrow 4\lambda \sin \theta =n\lambda$

$\Rightarrow \sin \theta =\frac{n}{4}$

Step 3: Find the possible values of n from the equation

The maximum value of $\sin \theta$ cannot exceed $1$.

$\because$ Possible values of $n$ are $0,\pm 1,\pm 2,\pm 3,\pm 4$

Thus, the maximum number of possible interference maxima is equal to $9$.