Q134 / 150 · MHT CET 2026 (19 April Shift 1) · Physics

In a single slit diffraction pattern, the distance between the plane of the slit and the screen is $1.4$ m. The width of the slit is $0.66$ mm. The second maximum is formed at the distance of $2.8$ mm from the center of the screen. The wavelength of light used is

  • A
    $6500 \ \text{Å}$
  • B
    $5600 \ \text{Å}$
  • Correct answer
    C
    $5280 \ \text{Å}$
  • D
    $4600 \ \text{Å}$

Solution

The position of the $n$-th secondary maximum in a single slit diffraction pattern is given by:

$y_n = \dfrac{(2n + 1)\lambda D}{2a}$

For the second maximum, $n = 2$:

$y_2 = \dfrac{5\lambda D}{2a}$

Rearranging for the wavelength $\lambda$:

$\lambda = \dfrac{2a y_2}{5D}$

Given values are:
$D = 1.4$ m
$a = 0.66$ mm $= 0.66 \times 10^{-3}$ m
$y_2 = 2.8$ mm $= 2.8 \times 10^{-3}$ m

Substituting the values into the formula:

$\lambda = \dfrac{2 \times 0.66 \times 10^{-3} \times 2.8 \times 10^{-3}}{5 \times 1.4}$

$\lambda = \dfrac{3.696 \times 10^{-6}}{7}$

$\lambda = 0.528 \times 10^{-6}$ m

$\lambda = 5280 \times 10^{-10}$ m $= 5280 \ \text{\AA}$

Answer: $5280 \ \text{Å}$

Hence, the correct answer is option: 'C'.

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