JEE Main 2025 (Online) 4th April Evening Shift | Wave Optics Question 3 | Physics

Two polarisers $P_1$ and $P_2$ are placed in such a way that the intensity of the transmitted light will be zero. A third polariser $P_3$ is inserted in between $P_1$ and $P_2$, at particular angle between $P_2$ and $P_3$. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers $P_2$ and $P_3$ is :

$\pi / 6$

$\frac{\pi}{3}$

$\frac{\pi}{4}$

$\pi / 8$

JEE Main 2025 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

In a Young's double slit experiment, the slits are separated by 0.2 mm . If the slits separation is increased to 0.4 mm , the percentage change of the fringe width is :

$25 \%$

$50 \%$

$0 \%$

$100 \%$

JEE Main 2025 (Online) 3rd April Evening Shift

MCQ (Single Correct Answer)

-1

Two monochromatic light beams have intensities in the ratio 1:9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

$8: 1$

$4: 1$

$3: 1$

$9: 1$

JEE Main 2025 (Online) 3rd April Evening Shift

MCQ (Single Correct Answer)

-1

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

$3: 1$

$(2 \sqrt{2}+1):(2 \sqrt{2}-1)$

$9: 1$

$(3+2 \sqrt{2}):(3-2 \sqrt{2})$

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