JEE Main 2025 (Online) 3rd April Evening Shift | Wave Optics Question 7 | Physics

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 :

$3: 1$

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

$9: 1$

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

JEE Main 2025 (Online) 3rd April Evening Shift

MCQ (Single Correct Answer)

-1

A monochromatic light of frequency $5 \times 10^{14} \mathrm{~Hz}$ travelling through air, is incident on a medium of refractive index ' 2 '. Wavelength of the refracted light will be :

400 nm

300 nm

600 nm

500 nm

JEE Main 2025 (Online) 2nd April Morning Shift

MCQ (Single Correct Answer)

-1

A light wave is propagating with plane wave fronts of the type $x+y+z=$ constant. Th angle made by the direction of wave propagation with the $x$-axis is :

$\cos ^{-1}(2 / 3)$

$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\cos ^{-1}\left(\frac{1}{3}\right)$

$\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$

JEE Main 2025 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

At the interface between two materials having refractive indices $n_1$ and $n_2$, the critical angle for reflection of an em wave is $\theta_{1C}$. The $\mathrm{n}_2$ material is replaced by another material having refractive index $n_3$ such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2 C}$. If $n_3>n_2>n_1 ; \frac{n_2}{n_3}=\frac{2}{5}$ and $\sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2}$, then $\theta_{1 C}$ is :

$ \sin^{-1}\left( \frac{1}{6n_1} \right) $

$ \sin^{-1}\left( \frac{1}{3n_1} \right) $

$ \sin^{-1}\left( \frac{5}{6n_1} \right) $

$ \sin^{-1}\left( \frac{2}{3n_1} \right) $

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