question_answer

What is diffraction of light? Draw and discuss the graph showing the variation of intensity with angle in a single slit diffraction experiment.

How would the diffraction pattern of a single slit be affected when

(i) the width of the slit is decreased?

(ii) the monochromatic source of light is replaced by a source of white light?

Or

(i) Derive the formula for refraction at concave refracting surface when the object lies in the rarer medium. Write the sign convention used.

(ii) What happens to the focal length of the lens when it is immersed in water.

           

Answer:

        Diffraction of Light The phenomenon of bending of light around the corners of an obstacle and spreading into the region of geometrical shadow is known as diffraction of light. Variation of intensity of light with diffraction angle \[\theta \] is shown graphically below: Diffraction angle \[\theta \to \]

The salient features of variation of intensity of light with diffraction angle \[\theta \] are as follows:

(a) The intensity of central maxima is greatest.

(b) The intensity of secondary maxima decrease with the increase of order of the maxima and are in the ratio

\[1:\frac{1}{21}:\frac{1}{61}:\frac{1}{121}...\] The angular width of secondary minima in the diffraction pattern of single slit are given by \[{{\theta }_{n}}=n\lambda /d\]                         ?(i) Where, \[\lambda \] is wavelength of monochromatic light, d is the slit width and n refers the order of secondary minima. (i) The diffraction pattern gets widened when slit width (d) decreases as angular diffraction \[({{\theta }_{n}})\]increases             \[{{\theta }_{n}}\propto \frac{1}{d}\] (ii) On using white light instead of monochromatic light in single slit, a coloured diffraction fringe pattern is obtained due to dispersion of light into constituting colours. But central maximum is white, consequently, fringe of higher wavelength (say red light) are wider than smaller wavelength (say blue light). Also, greater dispersion of light taken place corresponding to higher order diffraction pattern which leads to overlapping of different colours of VIBGYOR (bandwidth \[\propto \lambda \]). Or (i) Consider APB is a concave spherical surface of radius R separating the two medium of refractive indices \[{{\mu }_{1}}\] and \[{{\mu }_{2}}.\] Refraction at a concave surface when the object lies in the rarer In \[\Delta \,ONC,\] Exterior angle = Sum Of interior angles \[\gamma =i+\alpha \Rightarrow i=\gamma -\alpha \]                                ?(i) In \[\Delta \,INC,\] Exterior angle = Sum of interior angles \[\gamma =\beta +r\Rightarrow r=\gamma -\beta \]                                   ?(ii) Using Snell's law, \[\frac{\sin i}{\sin r}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\] \[\because \]Angles i and r are very small             \[\frac{i}{r}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\Rightarrow {{\mu }_{1}}i={{\mu }_{2}}r\]                  ?(iii) Putting the values of i and r from Eqs. (i) and (ii) into Eq. (iii) \[{{\mu }_{1}}(\gamma -\alpha )={{\mu }_{2}}(\gamma -\beta )\] \[{{\mu }_{2}}\beta -{{\mu }_{1}}\alpha =({{\mu }_{2}}-{{\mu }_{1}})\gamma \]                     ?(iv) \[\because \]       Aperture is very small. \[\therefore \]      \[M\to P\] In \[\Delta \,ONM,\]         \[\alpha \approx \tan \alpha =\frac{NM}{OM}\approx \frac{h}{PO}=\frac{h}{-u}\] \[\Delta \,INM,\]              \[\beta \approx \tan \beta =\frac{NM}{IM}\approx \frac{h}{PI}=\frac{h}{-v}\] \[\Delta \,CNM,\] \[\gamma \approx \tan \gamma =\frac{NM}{CM}\approx \frac{h}{PC}=\frac{h}{-R}\] Putting the values of \[\alpha ,\beta \] and \[\gamma ,\] in Eq. (iv), we get \[{{\mu }_{2}}\frac{h}{-v}-{{\mu }_{1}}\frac{h}{-\mu }=({{\mu }_{2}}-{{\mu }_{1}})\frac{h}{-R}\Rightarrow \frac{{{\mu }_{2}}}{v}-\frac{{{\mu }_{1}}}{u}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}\] Sign conventions (a) All distances are measured from optical centre. (b) Distance measured in the direction of incident ray of light are taken positive and vice-versa. (ii) Now as we know \[\frac{1}{f}=(\mu -1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] When a lens is immersed in water, refractive index \[(\mu )\] decreases. Hence, focal length increases.