# Wave optics- number of minimas.

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Attempt:

Let m be the integer associated with $420 nm$
and n be the integer associated with $540 nm$

$d \sin \theta = k \lambda$ $k\in Z$
Clearly,

$m_{max} = 180$

$n_{max} = 140$

$(2m+1) \lambda_1 = (2n+1)\lambda_2$ (condition for dark fringes to overlap)

$\implies \dfrac{2m+1}{2n+1} = \dfrac 97$

$\implies m = \dfrac{1+7n +2n }{7}$

Hence, we obtain, for m to be an integer: $2n+1 = 7k$

$\implies n = \dfrac{7k-1}{2}$ where $k \in Z$

Now note that k must be odd since odd-1 = even

Thus, using $n \le 140$, $k_{max} = 39$

Now, we have to consider only odd values of k which are $1,3,...39$ = 20 numbers

Thus, we have 20 minimas on the upper side and 20 on the lower,
Total minimas = $20+20 = 40$

But answer is $D$

Question 1: What is wrong with my method?

Question 2: Considering that this is a JEE Mains problem, what is the fastest 2 minute way to do it?