MHT CET 2026 19th April Morning Shift | Wave Optics Question 10 | Physics

In Young's double slit experiment, the wavelength of light used is $\lambda$. The intensity on the screen at a point for path difference '$\lambda$' is 'X'. The intensity at the point for path difference $\left(\dfrac{\lambda}{6}\right)$ is ($\cos 180^\circ = -1$, $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$)

$\dfrac{X}{6}$

$\dfrac{X}{2}$

$\dfrac{3X}{4}$

$\dfrac{4X}{3}$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

-0

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

$6500 \ \text{Å}$

$5600 \ \text{Å}$

$5280 \ \text{Å}$

$4600 \ \text{Å}$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

-0

In Young's double slit experiment, for the $n^{\text{th}}$ dark fringe ($n = 1, 2, 3, \ldots$) the phase difference of the interfering waves in radian will be

$n \cdot \dfrac{\pi}{2}$

$(2n + 1)\pi$

$(2n - 1)\pi$

$(2n - 1)\dfrac{\pi}{2}$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

A beam of light of intensity $I_0$ falls on a system of three polaroids which are arranged in succession such that the pass (transmission) axis is turned through $60^{\circ}$ with respect to preceding one. The fraction of the incident light intensity that passes through the system is $\left(\cos 60^{\circ}=\frac{1}{2}\right)$

$\frac{1}{8}$

$\frac{1}{32}$

$\frac{1}{16}$

$\frac{1}{2}$

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