Question

The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I₀. If one of the slits is closed, the intensity at this point will be
(a) I₀ (b) I₀/4 (c) I₀/2 (d) ​4I₀

Answer 1

dear student

$$\begin{gathered} ThemaximumintensityintheYoung\text{'}sdoubleslitexperimentisgivenby \\ {I}_{max}={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2 \\ letthismaaximumintensityis{I}_o \\ {I}_o={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2 \\ where{I}_{1}and{I}_{2}aretheintensitiesofthetwosources.Astheslitsareidenticaland \\ placedsymmetricallyfromsource \\ {I}_1=I_2 \\ hence \\ {I}_o={\left(\sqrt{I_1}+\sqrt{I_1}\right)}^2 \\ {\left(2\sqrt{I_1}\right)}^2=I_o \\ 4I_1=I_o \\ {I}_1=\frac{I_o}{4} \\ soifoneslitisclosedtheintensitywillbe\frac{I_o}{4}. \end{gathered}$$