Umakant Kondapure, Collin Fernandes, Nipun Bhatia, Vikram Bathula and, Ketki Deshpande Solutions for Chapter: Interference and Diffraction, Exercise 3: Competitive Thinking

In a Young's double-slit experiment, bi-chromatic light of wavelengths $400 \mathrm{nm}$ and $560 \mathrm{nm}$ are used. The distance between the slits is $0.1 \mathrm{mm}$ and the distance between the plane of the slits and the screen is $1 \mathrm{m}$. The minimum distance between two successive regions of complete darkness is

In Young's double slit experiment, intensity at a point is $\frac{1}{4}$ of the maximum intensity. Angular position of this point is 

In a double slit experiment, the two slits are $1 \mathrm{mm}$ apart and the screen is placed $1 \mathrm{m}$ away. A monochromatic light of wavelength $500 \mathrm{nm}$ is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern? 

In Young's double slit experiment the separation $d$ between the slits is $2 \mathrm{mm},$ the wavelength $\lambda$ of the light used is$5896 \mathrm{\AA }$ and distance $D$ between the screen and slits is $100 \mathrm{cm}.$ It is found that the angular width of the fringes is $0.20^{\circ} .$ To increase the fringe angular width to $0.21^{\circ}$ (with same $\lambda$ and $D$ ) the separation between the slits needs to be changed to

The angular width of the central maximum in a single slit diffraction pattern is $60^{\circ} .$ The width of the slit is $1 \mu \mathrm{m}$. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance $50 \mathrm{cm}$ from the slits. If the observed fringe width is $1 \mathrm{cm},$ what is slit separation distance? (i.e., distance between the centres of each slit.) 

The diameter of the pupil of human eye is $2.5 \mathrm{mm}$. Assuming the wavelength of light used is $5000 \mathrm{\AA }$. What must be the minimum distance between two point like objects to be seen clearly if they are a distance of $5 \mathrm{m}$ from the eye?

Assuming human pupil to have a radius of $0.25 \mathrm{cm}$ and a comfortable viewing distance of $25 \mathrm{cm},$ the minimum separation between two objects that human eye can resolve at $500 \mathrm{nm}$ wavelength is

The box of a pin hole camera, of length $L$ has a hole of radius $a$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $\left.b_{\min }\right)$ when