A ray of light goes from medium 1 to medium 2, velocity of light in the two mediums are \[{{C}_{1}}\] and \[{{C}_{2}},\] respectively. For an angle of incidence, \[\theta \] in medium 1. The corresponding angle of refraction in medium 2 is \[\theta /2.\] |
(i) Which of the two media is optically denser and why? |
(ii) Establish the relationship among \[\theta ,\] \[{{c}_{1}}\] and \[{{c}_{2}}.\] |
(iii) The critical angle of incidence in a glass slab placed in air \[45{}^\circ .\] What will be the critical angle when it is immersed in water of refractive index 1.33? |
Or |
A beam of light of wavelength 400 nm is incident normally on a right angled prism as shown in the figure at night. |
It is observed that the light just grazes along the surface AC after falling on it. |
Given that the refractive index of the material of the prism varies with the wavelength \[\lambda \] as per the relation, |
Answer:
(i) Smaller value of angle of refraction \[\left( \frac{\theta }{2} \right)\] as compared to angle of incidence \[(\theta )\] indicates bending of light towards the normal in second medium. Hence, medium 2 is optically denser. (ii) By Snell?s law, \[{}_{1}{{\mu }_{2}}=\frac{\sin i}{\sin r}=\frac{\sin \theta }{\sin \theta /2}\] \[=\frac{2\sin \theta /2\cos \theta /2}{\sin \theta /2}=2\cos \theta /2\] Also, \[{}_{1}{{\mu }_{2}}=\frac{{{c}_{1}}}{{{c}_{2}}}\] \[\therefore \] \[\frac{{{c}_{1}}}{{{c}_{2}}}=2\cos \theta /2\] or \[\theta =2{{\cos }^{-1}}\left( \frac{{{c}_{1}}}{2{{c}_{2}}} \right)\] (iii) \[{}^{a}{{\mu }_{g}}=\frac{1}{\sin {{i}_{c}}}=\frac{1}{\sin 45{}^\circ }=\sqrt{2}=1.414\] Refractive index of glass w.r.t. water will be \[{}^{w}{{\mu }_{g}}=\frac{{}^{a}{{\mu }_{g}}}{{}^{a}{{\mu }_{w}}}=\frac{1.414}{1.33}\] When glass slab is immersed in water, then critical angle \[i{{'}_{c}}\] is given by \[\sin i{{'}_{c}}=\frac{1}{{}^{w}{{\mu }_{g}}}=\frac{1}{\frac{1.414}{1.33}}=\frac{1.33}{1.414}=0.9432\]\[\therefore i{{'}_{c}}=70{}^\circ .36'\] Or Wavelength of incident light, \[\lambda =400\,nm=400\times {{10}^{-9}}m=4\times {{10}^{-7}}m\] For refracting surface AB, \[{{i}_{1}}=0{}^\circ ,\] \[{{r}_{1}}=0{}^\circ \] For refracting surface AC, \[{{r}_{2}}=90{}^\circ \] and \[{{i}_{2}}=\theta \] Using Snell's law at surface AC, \[\frac{\sin {{i}_{2}}}{{{\operatorname{sinr}}_{2}}}=\frac{1}{\mu }\Rightarrow \frac{\sin \theta }{\sin 90{}^\circ }=\frac{1}{\mu }\] \[\Rightarrow \] \[\frac{1}{\sin \theta }=\mu \] \[[\because \sin 90{}^\circ =1]\] \[\Rightarrow \] \[\mu =\frac{1}{0.625}=1.6\] \[[\text{given},\sin \theta =0.625]\] But the relation is \[\mu =12+\frac{b}{{{\lambda }^{2}}}\] [given] \[\therefore \] \[1.6=12+\frac{b}{{{(4\times {{10}^{-7}})}^{2}}}\] \[b=0.4\times {{(4\times {{10}^{-7}})}^{2}}=6.4\times {{10}^{-14}}{{m}^{2}}\] For wavelength, \[\lambda =5000\,\overset{{}^\circ }{\mathop{A}}\,=5\times {{10}^{-7}}m\] Refractive index, \[\mu =1.2+\frac{6.4\times {{10}^{-14}}}{{{(5\times {{10}^{-7}})}^{2}}}=1.2+\frac{6.4}{25}\] \[=1.2+0.256=1.456\]
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