MHT CET · Physics · Wave Optics

In Young’s double slit experiment fifth dark fringe is formed opposite to one of the slit. If D is the distance between the slits and the screen and d is the separation between the slits, then the wavelength of light used is

A $\frac{d^{2}}{5 D}$
B $\frac{d^{2}}{9 D}$
C $\frac{d^{2}}{6 D}$
D $\frac{d^{2}}{15 D}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{d^{2}}{9 D}$

Step-by-step Solution

Detailed explanation

Position of nth dark fringes in Young’s double slit experiment is given,

x_{n} = \frac{D}{d} (2 n - 1) \frac{λ}{2}

\Rightarrow λ = \frac{2 x_{n} d}{D (2 n - 1)}

… (i)
Since, fifth dark fringe is formed opposite to one of the slit,

∴ 2 x_{5} = d

(distance between slits)
Put

n = 5

in Eq. (i), we get

λ = \frac{2 x_{5} d}{D (2 \times 5 - 1)}

= \frac{d . d}{D (10 - 1)} [s i n c e, 2 x_{5} = d]

\frac{d^{2}}{9 D}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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