List of top Physics Questions on Wave Optics asked in MHT CET

In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index '\(\mu\)' is placed in one of the interfering paths. Now \(7^{\text{th}}\) bright fringe is obtained at the same point. If '\(\lambda\)' is the wavelength of light used, the thickness of film is equal to
When a ray of light is refracted from one medium to another, then the wavelength changes from \(6000\text{ \AA}\) to \(4000\text{ \AA}\). The critical angle for the interface will be
A plane wavefront of width ' \( x \) ' is incident on an air-water interface and the corresponding refracted wavefront has a width ' \( y \) ' as shown in figure. The refractive index of air with respect to water in terms of distances ' \( w \) ' and ' \( z \) ' is (AD = w, CB = z)

A double slit experiment is immersed in water of refractive index 1.33. The slit separation is 1 mm and the distance between slit and screen is 1.33 m. The slits are illuminated by a light of wavelength \(6300 \, \text{Å}\). The fringe width is
In biprism experiment, 21 fringes are observed in a given region using light of wavelength 4800 Å. If light of wavelength 5600 Å is used, the number of fringes in the same region will be
Two coherent sources 'P' and 'Q' produce interference at point 'A' on the screen, where there is a dark band which is formed between 4th and 5th bright band. Wavelength of light used is $6000\text{ \r{A}}$. The path difference PA and QA is
Two coherent sources of wavelength \(\lambda\) produce steady interference pattern. The path difference corresponding to \(10^{\text{th}}\) order maximum will be
In a biprism experiment, fifth dark fringe is obtained at a point. A thin transparent film of refractive index '\(\mu\)' is placed in one of the interfering paths. Now \(7^{\text{th}}\) bright fringe is obtained at the same point. If '\(\lambda\)' is the wavelength of light used, the thickness of film is equal to
When a ray of light is refracted from one medium to another, then the wavelength changes from \(6000\text{ \AA}\) to \(4000\text{ \AA}\). The critical angle for the interface will be
In Young's double slit experiment, the $8^{\text{th}}$ maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum and the $6^{\text{th}}$ maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $\frac{d_2}{d_1}$ is