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

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?

  1. $12.5 cm$

  2. $12 cm$

  3. $13 cm$

  4. $14 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

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

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

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

  4. $6 \, \times \, 10^{-3} \, rad$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

$\therefore \, \theta \, \cong \, sin \, \theta \, = \, 2 \, \times \, 10^{-3} \, rad$
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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:

  1. 12.13 m

  2. 6.67 m

  3. 3.33 m

  4. 2.19 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


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

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

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

  1. 22 m

  2. 32 m

  3. 42 m

  4. 52 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:
 Aperture Width $a =4 $mm
Fresnel distance, $Z _F \, = \, \dfrac{a^2}{\lambda} \, = \, \dfrac{(4 \, \times \, 10^{-3})^2}{500 \, \times \, 10^{-9}}$
$\therefore \, z _p \, = \, 32 \, m$
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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 $\theta$ =

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

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

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

  4. $sin^{-1} [\frac{\sqrt{3}}{2}]$
Reveal answer Fill a bubble to check yourself
C Correct answer
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. $sin \theta _n = \dfrac{3 \lambda}{2d} = \dfrac{ 3 \times 5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}} = sin \theta _n = \dfrac{3}{4} or \theta _n = sin^{-1}(3/4)$

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

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

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

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

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

  4. $\dfrac{a _2}{a _1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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

  1. colored

  2. white

  3. dark

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.