question_answer

    In Youngs double slit experiment, the slits are 3 mm apart. The wavelength of light used is \[5000\text{ }\overset{o}{\mathop{\text{A}}}\,\]and the distance between the slits and the screen is 90 cm. The fringe width in mm is

A)  1.5                                        

B)  0.015

C)  2.0                                        

D)  0.15

Correct Answer: D

Solution :

                Let\[\lambda \]be wavelength of monochromatic light, used to illuminate the slit 5, and d be the distance between coherent sources, then width of slits is given by where D is distance between screen and source. Given,\[d=3\text{ }mm,\text{ }\lambda =5000\text{ }\overset{o}{\mathop{\text{A}}}\,=5\times {{10}^{-7}}m\]                                 \[=5\times {{10}^{-4}}mm,\] \[D=90\text{ }cm=900\text{ }mm.\] \[\therefore \]\[W=\frac{5\times {{10}^{-4}}\times 900}{3}=15\times {{10}^{-2}}mm\] \[=0.15\text{ }mm\]