In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is 2.4 cm. The frequency of light used is ______________ $\times$ 1014 Hz.
Answer (integer)
5
Solution
8$\beta$ = 2.4 cm<br><br>${{8\lambda \Delta } \over d}$ = 2.4 cm<br><br>$${{8 \times 1.5 \times c} \over {0.3 \times {{10}^{ - 3}} \times f}} = 2.4 \times {10^{ - 2}}$$<br><br>f = 5 $\times$ 10<sup>14</sup> Hz
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
This question is part of PrepWiser's free JEE Main question bank. 197 more solved questions on Optics are available — start with the harder ones if your accuracy is >70%.