The maximum number of possible interference m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The condition for interference maxima in Young's double-slit experiment is given by $\Delta x = d \sin 	heta = n \lambda$,where $n$ is an integer

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The condition for interference maxima in Young's double-slit experiment is given by $\Delta x = d \sin 	heta = n \lambda$,where $n$ is an integer.Given that the slit separation $d = 2 \lambda$,we substitute this into the equation:$2 \lambda \sin 	heta = n \lambda$$2 \sin 	heta = n$$\sin 	heta = \frac{n}{2}$Since the maximum value of $\sin 	heta$ is $1$,we have $-1 \le \frac{n}{2} \le 1$,which implies $-2 \le n \le 2$.The possible integer values for $n$ are $-2, -1, 0, 1, 2$.Counting these values,we get a total of $5$ interference maxima.

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

Similar Questions

If $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$,then $A+C-B$ is equal to :

Which of the following compounds will exhibit geometrical isomerism?

$A$ screen is placed $50 \, cm$ from a single slit, which is illuminated with $6000 \, \mathring{A}$ light. If the distance between the first and third minima in the diffraction pattern is $3 \, mm$, the width of the slit is.....$mm$

Haploids are more suitable for mutation studies than diploids. This is because

In which of the following arrangements is the order $NOT$ according to the property indicated against it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module