Wave Optics · Physics · MHT CET (Biology)

Monochromatic light of wavelength $4800 \mathop {\rm{A}}\limits^{\rm{o}}$ falls on slit of width ' $a$ '. For what value of ' $a$ ', the first maximum falls at $30^{\circ} .\left(\sin 30^{\circ}=0.5\right)$

In Young's double slit experiment, waves from slits ' $\mathrm{S}_1$ ' and ' $\mathrm{S}_2$ ' have a path difference of $\frac{\lambda}{4}$ and $\frac{\lambda}{6}$ respectively at points $x$ and $y$ on the screen. The ratio of intensities at points $y$ to that at $x$ is $\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 30^{\circ}=\sqrt{3} / 2,\right)$

' n ' polarising sheets are arranged such that each makes an angle $45^{\circ}$ with the proceeding sheet. An unpolarised light of intensity I is incident into this arrangement. The output intensity is I/ 64 . The value of $n$ will be $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)$

In an interference experiment, the phase difference between the waves reaching a first dark point is

In Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ are $\mathrm{I}_1$ and $\mathrm{I}_2$ respectively. If $\mathrm{I}_0$ denotes the intensity produced by each one of the individual slits then the ratio $\left(\mathrm{I}_1+\mathrm{I}_2\right): \mathrm{I}_0$ is $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)\left(\cos 60^{\circ}=0.5\right)$

In Young's double slit experiment, fringe width is 1.4 mm with light of wavelength $6000 $$\mathop {\rm{A}}\limits^{\rm{o}} $. If the light of wavelength $5400 $$\mathop {\rm{A}}\limits^{\rm{o}} $ is used, with no other change in the experimental set up. The change in fringe width is

In a single slit diffraction experiment, slit of width ' a ' and incident light of wavelength $5600 $$\mathop {\rm{A}}\limits^{\rm{o}} $, the first minimum is observed at angle $30^{\circ}$. The first secondary maximum is observed at angle $\left(\operatorname{Sin} 30^{\circ}=0.5\right)$

In biprism experiment the maximum intensity is ' $\mathrm{I}_0$ '. If the path difference between the two interfering waves is ' $\lambda / 4$ ' then intensity at the point on the screen is

$$ \left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right] $$

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