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QJEEM21Ph95fa: JEE Main 2021 Physics - Class 12 JEEM21Ph95fa Solution: White light is passed through a double slit and interference is …

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.

Step-by-Step Verified Solution

Position of bright fringe y = n Dλd{{D\lambda } \over d}

y1 of red = Dλrd{{D{\lambda _r}} \over d} = 3.5 mm

λ\lambda r = 3.5 ×\times 10 −- 3 dD{d \over D}

Similarly, λ\lambda v = 2 ×\times 10 −- 3 dD{d \over D}

λr−λv=(1.5×10−3)(0.3×10−31.5){\lambda _r} - {\lambda _v} = (1.5 \times {10^{ - 3}})\left( {{{0.3 \times {{10}^{ - 3}}} \over {1.5}}} \right)

= 3 ×\times 10 −- 7 = 300 nm

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