White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. the difference in wavelengths of red and violet light is ................ nm.
Answer (integer)
300
Solution
Position of bright fringe y = n${{D\lambda } \over d}$<br><br>y<sub>1</sub> of red = ${{D{\lambda _r}} \over d}$ = 3.5 mm<br><br>$\lambda$<sub>r</sub> = 3.5 $\times$ 10<sup>$-$3</sup> ${d \over D}$<br><br>Similarly, $\lambda$<sub>v</sub> = 2 $\times$ 10<sup>$-$3</sup> ${d \over D}$<br><br>$${\lambda _r} - {\lambda _v} = (1.5 \times {10^{ - 3}})\left( {{{0.3 \times {{10}^{ - 3}}} \over {1.5}}} \right)$$<br><br>= 3 $\times$ 10<sup>$-$7</sup> = 300 nm
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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