Solution:

Ans. The phenomenon observed is diffraction.

(i) At the cental maxima : The contributions due to the secondary wavelets, from all parts of the wave front (at the slit), arrive in phase at the central maxima $\theta =0$

At the secondary maxima :

It is only the contributions from (Dearly) $1/3$ (or $1/5$, or $1/7,\dots$ ) of the secondary maxima. These occur at points for which

$\theta \cong (n+\frac{1}{2})\frac{\lambda }{a}(n=0,1,2,3,\dots )$

1

At the minima :

The contribution, from ‘corresponding pairs’, of the sub-parts of the incident wavefront, cancel each other and the net contribution, at the location of the minima, is zero. The minima occur at points for which, $\theta =n\frac{\lambda }{a}(n=1,2,3,\dots )$ [Note: Award these $(1+1+1)$ marks if the student draws the diagram and writes the conditions, for $\theta$, for the three cases.]

(ii) There is a significant fall in intensity at the secondary maxima because the intensity there, is only due to the contribution of (nearly)( $1/3$ or $1/5$ or $1/7,\dots ..$.$)oftheincidentwavefronts.$

(iii) The size of the central maxima would get halved when width of the slit is doubled.

1

[CBSE Marking Scheme 2016]

AT Q. 13. (i) (a) ‘Two independent monochromatic sources of light cannot produce a sustained interference pattern’. Give reason.

(b) Light waves each of amplitude " $a$ " and frequency " $\omega$ “, emanating from two coherent light sources superimpose at a point. If the displacements due to these waves are given by $y_1=a\cos \omega t$ and $y_2=a\cos (\omega t+\varphi )$, where $\varphi$ is the phase difference between the two, obtain the expression for the resultant intensity at the point.

(ii) In Young’s double slit experiment, using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is $K$ units. Find out the intensity of light at a point where path difference is $\lambda /3$.

[Delhi I, II, III 2014 ]

Ans. (i) Try yourself similar to Q. 7 SATQ-II

(ii) A path difference of $\lambda$, corresponds to a phase difference of $2\pi$.

$\therefore$ The intensity, $I=4a^2\Rightarrow a^2=\frac{I}{4}$

$1/2$

A path difference of $\frac{\lambda }{3}$, corresponds to a phase difference of $\frac{2\pi }{3}$.

$1/2$

$\therefore$ Intensity $=4a^2{\cos }^2\varphi /2$

$$\begin{array}{rlr}& =4\times a^2\times {\cos }^2\frac{2\pi /3}{2}& =4\times \frac{I}{4}\times {\left(\frac{1}{2}\right)}^2=\frac{I}{4}\end{array}$$

[CBSE Marking Scheme 2014]

TOPIC-3

Resolving Power of Optical instruments and Polarisation of Light

Revision Notes

Resolving Power of Optical Instruments : The ability of an optical instrument to separate small or closely adjacent images.

Border line resolving is the required minimum angle $\theta$ between two objects so that they may be resolved.

Rayleigh criteria for border line resolving

If opening is circular (which in case of most optical instruments), Rayleigh criteria of border line resolving is

$$\mathrm{\Delta }\theta =\frac{1.22\lambda }{a}=\frac{D}{d}$$

Resolving Power of Microscope

$RP=\frac{1}{\text{ limit of resolution }}$

Limit of resolution of microscope

$$d_{min}=\frac{1,22\lambda }{2\sin \beta }$$

And $P\propto \frac{1}{d_{min}}$ (the minimum distance it may resolve, the more powerful is the microscope). Hence we may reduce the value of $\lambda$ by filling the liquid of refractive index $n$ around the objective lens. In this case the medium between the object and the objective lens gotair but a medium of refractive index $n$.

$>$ Resolving Power of Telescope

$$d_{min}=\frac{1.22\lambda }{2n\sin \beta }$$

Hence to differentiate minimum angular resolution, aperture ’ $a$ ’ of the objective lens should be as large as possible.

$$\mathrm{RP}=\frac{a}{1.22\lambda }$$

Fresnel’s distance limit of ray optics

$$Z=\frac{a^2}{\lambda }$$

where, $a$ is the size of aperture.

Doppler’s effect

When there is relative motion between source of light and observer, the frequency as apparent to observer is different and there is change, according to the following relation.

$$\frac{\mathrm{\Delta }v}{v}=\frac{-v_{\text{relative }}}{c}$$

where, $v=$ original frequency

$\mathrm{\Delta }v=$ change in frequency

$v_{\text{relative }}=$ relative velocity. It is considered to be positive when source is moving away from the observer.

If there is redshift in doppler effect then source is moving away and if there is blue shift in doppler effect then source is moving towards observer.

Polarisation

In an unpolarized wave, the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation.

Hence in unpolarized light wave, vibrations of electric fields are in more than one direction. Polarized light waves are light waves in which the vibrations occur in a single plane.

The process of transforming unpolarized light into polarized light is known as polarisation.

• Light is unpolarized by nature but it may be converted into fully or partial polarized light

The light having oscillation only in one plane is called polarised or plane polarised.

• The plane perpendicular to the plane of oscillation is called plane of polarisation.

Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.

Polaroids are thin films of ultramicroscopic, crystals of quinine iodo sulphate with their optical axes parallel to each other.

Polaroids allow the light oscillations parallel to the transmission axis to pass through them.

• If an unpolarised light is incident on a polaroid, the transmitted light is plane polarised as shown below :

Here, the vertical oscillations are transmitted because the transmission axis is also vertical. The horizontal oscillations are not transmitted. That is why, on the right hand side there are nodots, at the intersection of lines.

• When polaroids are used to convert unpolarised light into polarised they areknown as polarisers.
• When polaroids are used to detect polarised light, they are known as analyse
• If the transmission axes of the polariser and analyser are parallel, then whole of the polarised light passes through the analyser.

If the transmission axis of the analyser is perpendicular to that of polariser, then no light passes through the analyser. Such polariser and analyser are said to be crossed.

• Malus’ law : It states about thesintensity of a polarised light when it is altered to pass through a polariser at an angle $\theta$. According to this law tf ${0}$betheintensityofthepolarisedlightincidentontheanalyserand$\theta$betheanglebetweenthetransmissionaxesofthepolariserandanalyser,thentheintensityofthelighttransmittedthroughtheanalyserisgivenby$I=I{0} \cos ^{2} \theta . “$

$$\text{ Percentage of polarisation }=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}\times 100$$

• Polarisation by reflection : When unpolarised light is incident on the boundary between two transparent media such that reflected wave is perpendicular to the refracted wave, then the reflected wave is a totally polarised wave. The angle of incidence in this case is called Brewster’s angle and is denoted by $i_B$. If refractive index of second medium is $\mu$ then

$$\begin{array}{rlrl}\mu & =\frac{\sin i}{\sin r}\mu & =\frac{\sin i_B}{\sin (90-i_B)}\mu & =\frac{\sin i_B}{\cos i_B}=\tan i_B\end{array}$$

This is known as Brewster’s law.

Polarisation by scattering : When scattered light coming to the observer such that source is at ${90}^{\circ }$, then scattered light coming to observer is polarised.

Applications of Polarisation :

Polaroids are used to reduce the intensity in window pane, sunglasses etc.

Polaroids are also used in photographic cameras and 3D movie cameras.

Used in making holograms.

Polarisation proves that light is transverse wave.

Chemical analysis of molecules.

Know the Formulae

Resolving Power of Microscope

Resolving Power of Telescope

$$RP=\frac{2n\sin \beta }{1.22\lambda }$$

Polarisation

• Percentage to polarisation is given by

$$\frac{I_{max}-I_{min}}{I_{max}+I_{min}}\times 100$$

$$\mu =\tan i_B$$

fObjective Type Question