Question
A double slit, $${S_1} - {S_2}$$ is illuminated by a light source $$S$$ emitting light of wavelength $$\lambda .$$ The slits are separated by a distance $$d.$$ A plane mirror is placed at a distance $$D$$ in front of the slits and a screen is placed at a distance $$2D$$ behind the slits. The screen receives light reflected only by the plane mirror. The fringe-width of the interference pattern on the screen is
A double slit, $${S_1} - {S_2}$$ is illuminated by a light source $$S$$ emitting light of wavelength $$\lambda .$$ The slits are separated by a distance $$d.$$ A plane mirror is placed at a distance $$D$$ in front of the slits and a screen is placed at a distance $$2D$$ behind the slits. The screen receives light reflected only by the plane mirror. The fringe-width of the interference pattern on the screen is
A.
$$\frac{{D\lambda }}{d}$$
B.
$$\frac{{2D\lambda }}{d}$$
C.
$$\frac{{3D\lambda }}{d}$$
D.
$$\frac{{4D\lambda }}{d}$$
Answer :
$$\frac{{4D\lambda }}{d}$$
Solution :
Due to reflection virtual source will be formed at distance $$D$$ from mirror.
The effective distance of the screen $$= 2D + 2D = 4D$$
∴ Fringe width $$ = \frac{{4D\lambda }}{d}$$
Due to reflection virtual source will be formed at distance $$D$$ from mirror.
The effective distance of the screen $$= 2D + 2D = 4D$$
∴ Fringe width $$ = \frac{{4D\lambda }}{d}$$