COMEDK Exam 2018 Solved Paper

(*) In double slit experiment, for possible interference maxima on the screen is given as
dsin‌θ=nλ
Given, slit width, d=4λ
‌∴‌‌4λsin‌θ=nλ
‌⇒‌‌4sin‌θ=n
‌Since, maximum value of sin‌θ=1
∴ For maximum number of possible interference maxima,
‌4⋅1=n
‌⇒‌‌n=4
‌‌ i.e., ‌‌‌n=0,1,2,3
∴ Maximum number of possible maxima will be 7 i.e. 0,±1,±2,±3
Hence, no option is correct.