Two sources S1 and S2 emitting light of wavelength 600 nm are placed at a distance 1.0 × 10-2 cm apart. A detector can be moved on the line S1 P which is perpendicular to S1 S2. The position of the farthest minimum detected is approximately

Question

Two sources S1 and S2 emitting light of wavelength 600 nm are placed at a distance 1.0 × 10-2 cm apart. A detector can be moved on the line S1 P which is perpendicular to S1 S2. The position of the farthest minimum detected is approximately (A) 1.5 m (B) 1.0 m (C) 1.07 m (D) 1.03 m

Tardigrade Admin · Accepted Answer

The position of farthest minimum detection occurs when the path difference is least and odd multiple of

\frac{λ}{2}

, i.e., condition for destructive interference and approaches zero as

P

moves to infinity.

So, if

S_{2} P = D

S_{1} P - S_{2} P = (2 n + 1) \frac{λ}{2}

for destructive interference.

(n = 0, 1, 2, \dots)

. For farthest distance
so

S_{1} P - S_{2} P = \frac{λ}{2}

\Rightarrow D^{2} + d^{2} - D = \frac{λ}{2}

\Rightarrow D^{2} + d^{2} = (D + \frac{λ}{2})^{2}

\Rightarrow d^{2} = D λ + \frac{λ ^{2}}{4}

D = \frac{d ^{2}}{λ} - \frac{λ}{4}

= \frac{( 1.0 \times 1 0 ^{- 4} m ) ^{2}}{( 600 \times 1 0 ^{- 9} m )} - 150 \times 1 0^{- 9} m

= 107 c m

\Rightarrow D = 1.07 m

Q. Two sources S1​ and S2​ emitting light of wavelength 600nm are placed at a distance 1.0×10−2cm apart. A detector can be moved on the line S1​P which is perpendicular to S1​S2​. The position of the farthest minimum detected is approximately

1.5 m

1.0 m

1.07 m

1.03 m

Q. Two sources $S_{1}$ and $S_{2}$ emitting light of wavelength $600 nm$ are placed at a distance $1.0 \times 1 0^{- 2} c m$ apart. A detector can be moved on the line $S_{1} P$ which is perpendicular to $S_{1} S_{2}$ . The position of the farthest minimum detected is approximately