Question

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Youngs double-slit experiment, is

• A. infinite
• B. five
• C. three
• D. zero

Answer 1

Answer: (B)

five

Answer 2

For possible interference maxima on the screen, the condition is $d\sin \theta =n\lambda$ ... (i) Given: $d=slit-width=2\lambda$ $\therefore$ $2\lambda \sin \theta =n\lambda$ $\Rightarrow$ $2\sin \theta =n$ The maximum value of $\sin \theta$ is $1$ , hence, $n=2\times 1=2$ Thus, E (i) must be satisfied by 5 integer values $ie,$ $-2,\,\,-1,\,\,0,\,\,1,\,\,2$ . Hence, the maximum number of possible interference maxima is $5$ .