Question
In a double slit experiment slits $${S_1},{S_2}$$ is illuminated by a coherent light of wavelength $$\lambda .$$ The slits are separated by a distance $$d.$$ The experimental set up is modified by using plane mirrors as shown in figure. Find the fringe width of interference pattern on the screen.
In a double slit experiment slits $${S_1},{S_2}$$ is illuminated by a coherent light of wavelength $$\lambda .$$ The slits are separated by a distance $$d.$$ The experimental set up is modified by using plane mirrors as shown in figure. Find the fringe width of interference pattern on the screen.
A.
$$\frac{{\left( {3{D_1} + 2{D_2}} \right)\lambda }}{d}$$
B.
$$\frac{{\left( {2{D_1} + 3{D_2}} \right)\lambda }}{d}$$
C.
$$\frac{{\left( {3{D_2} - 3{D_1}} \right)\lambda }}{d}$$
D.
$$\frac{{\left( {3{D_1} - 2{D_2}} \right)\lambda }}{{2d}}$$
Answer :
$$\frac{{\left( {3{D_1} + 2{D_2}} \right)\lambda }}{d}$$
Solution :
The fringe width $$\beta $$ is given by, $$\beta = \frac{{D\lambda }}{d}$$
$$\eqalign{ & {\text{where}}\,\,D = {D_1} + \left( {{D_1} + {D_2}} \right) + \left( {{D_1} + {D_2}} \right) = 3{D_1} + 2{D_2} \cr & \therefore \beta = \frac{{\left( {3{D_1} + 2{D_2}} \right)\lambda }}{d}. \cr} $$
The fringe width $$\beta $$ is given by, $$\beta = \frac{{D\lambda }}{d}$$
$$\eqalign{ & {\text{where}}\,\,D = {D_1} + \left( {{D_1} + {D_2}} \right) + \left( {{D_1} + {D_2}} \right) = 3{D_1} + 2{D_2} \cr & \therefore \beta = \frac{{\left( {3{D_1} + 2{D_2}} \right)\lambda }}{d}. \cr} $$