Tag: physics

Questions Related to physics

In a Fresnel biprism experiment, the two positions of lens give separation between the slits as $16 $cm and$9 $cm, respectively. What is the actual distance of separation?

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $12.5 cm$

  2. $12 cm$

  3. $13 cm$

  4. $14 cm$

Show answer
Correct Option: B
Explanation:

A Fresnel Biprism is the variation on the Young's Slits experiment. The Fresnel biprism has two thin prisms which are joint at their bases to form an isosceles triangle. A single wavefront impinges on both prisms.

Separations between the slits,

$d _{1}= 16 cm$

and $d _{2}= 9 cm.$

Actual distance of separation (d)  can be computed with the formula as given:

$d = \sqrt {d _{1} \times d _{2}}\\$

$d = \sqrt {16 \times 9}\\$

$d = \sqrt {144} \\$

$d = 12 cm$

Thus actual distance will be $12 cm$.

Option B is correct.

A parallel beam of light of wavelength 6000 $\overset{0}{A}$ gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $2 \, \times \, 10^{-3} \, rad$

  2. $3 \, \times \, 10^{-3} \, rad$

  3. $1.8 \, \times \, 10^{-3} \, rad$

  4. $6 \, \times \, 10^{-3} \, rad$

Show answer
Correct Option: A
Explanation:
Here, $\lambda \, = \, 6000 \, \overset{0}{A} \, = \, 6000 \, \times \, 10^{-10} \, m \, = \, 6 \, \times \, 10^{-7} \, m \, a \, - \, 0.3 \, mm \, = \, 0.3 \, \times \, 10^{-4} \, m \, = \, 3 \, \times \, 10^{-4} \, m$

For first minima, $a sin \theta \, = \, \lambda$ where a is the slit widht 
$sin \, \theta = \, \dfrac{\lambda}{a} \, = \, \dfrac{6 \, \times \, 10^{-7}}{3 \, \times \, 10^{-4}} \, = \, 2 \, \times \, 10^{-3}$

As $sin \theta$ is very small

$\therefore \, \theta \, \cong \, sin \, \theta \, = \, 2 \, \times \, 10^{-3} \, rad$

Light of wavelength 600 nm is incident on an aperture of size 2 mm. The distance upto which light can travel such that its spread is less than the size of the aperture is:

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 12.13 m

  2. 6.67 m

  3. 3.33 m

  4. 2.19 m

Show answer
Correct Option: B
Explanation:


Given:
 Aperture Width, $a =2mm=2\times 10^{-3}mm$
The distance upto which light can travel is Fresnel distance, $Z _F \, = \, \dfrac{a^2}{\lambda} \, = \, \dfrac{(2 \, \times \, 10^{-3})^2}{600 \, \times \, 10^{-9}} \, = \, 6.67 \, m$

For what distance is ray optics a good approximation when the aperture is 4 mm wide and the wavelength is 500 nm? 

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 22 m

  2. 32 m

  3. 42 m

  4. 52 m

Show answer
Correct Option: B
Explanation:

Given:
 Aperture Width $a =4 $mm
Fresnel distance, $Z _F \, = \, \dfrac{a^2}{\lambda} \, = \, \dfrac{(4 \, \times \, 10^{-3})^2}{500 \, \times \, 10^{-9}}$
$\therefore \, z _p \, = \, 32 \, m$

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength $5500 A^0$, the first minimum is observed at angle $30^0$. The first secondary maximum is observed at an angle $\theta$ =

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $sin^{-1} [\frac{1}{\sqrt{2}}]$

  2. $sin^{-1} [\frac{1}{4}]$

  3. $sin^{-1} [\frac{3}{4}]$

  4. $sin^{-1} [\frac{\sqrt{3}}{2}]$

Show answer
Correct Option: C
Explanation:

Slit width = d
$\lambda = 5500 A^0 = 5.5 \times 10^{-7}m, \theta _n = 30^0$
For first secondary minima, $d sin \theta _n = \lambda$
$ d = \dfrac{\lambda}{sin \theta _n} = \dfrac{5.5 \times 10^{-7}}{sin 30^0} = 11 \times 10^{-7}$m
For the first secondary maxima, $d sin \theta _n = \dfrac{3 \lambda}{2}$
i.e. $sin \theta _n = \dfrac{3 \lambda}{2d} = \dfrac{ 3 \times 5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}} = sin \theta _n = \dfrac{3}{4} or \theta _n = sin^{-1}(3/4)$

The intensity of light at a distance $r$ from the axis of a long cylindrical source is inversely proportional to $r$.

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$I=\cfrac { { I } _{ o } }{ r } \ \therefore I\propto \cfrac { 1 }{ r } $

Intensity is inversely proportional to $r$.

In diffraction occurs through a single slit then intensity of first secondary maxima become ...................  % of central maxima :-

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $4$ %

  2. $25$ %

  3. $75$ %

  4. $50$ %

Show answer
Correct Option: C

Two slits of width $a _1$  and $a _2$ are illuminated by light of same wavelength.The first diffraction minima produced by each of them are in directions inclined at angles $\theta _1$ and $\theta _2$. The ratio of $sin\, \theta _1$ to $sin\,\theta _2$ is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $\dfrac{a _1}{a _2}$

  2. $\sqrt{\dfrac{a _1}{a _2}}$

  3. $\sqrt{\dfrac{a _2}{a _1}}$

  4. $\dfrac{a _2}{a _1}$

Show answer
Correct Option: A

When monochromatic light is replaced by white light in Fresnel's biprism arrangement, the central fringe is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. colored

  2. white

  3. dark

  4. None of these

Show answer
Correct Option: B
Explanation:

When white light refracts from the biprism, all colours would form coherent sources at different positions according to their refractive indices. However the position of their central fringe would be same, and hence they would all combine at that position to again give a white fringe.

Among the Fresnel zones the operative zones contributing intensity are :

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. last zones

  2. first few zones

  3. middle zones

  4. all the zones

Show answer
Correct Option: D
Explanation:

All the fresnel zones contribute to the intensity.