Question

In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

9.75 mm

15.6 mm

1.56 mm

7.8 mm

Solution

7.8 mm

For common maxima
n₁λ₁ = n₂λ₂
n₁ × 650 = n₂ × 520
$straight n subscript 1 over straight n subscript 2 space equals space 4 over 5 yd over straight D space equals space nλ straight y space equals fraction numerator 4 space straight x space 650 space straight x space 10 to the power of negative 9 end exponent space straight x space 1.5 over denominator 0.5 space straight x space 10 to the power of negative 3 end exponent end fraction space space straight y space equals space 7.8 space mm$

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