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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    A glass slab consists of thin uniform layers of  progressively decreasing refractive indices \[\text{RI}\] (see figure) such that the \[\text{RI}\]of any layer is \[\mu -m\,\Delta \mu .\]Here, \[\mu \]and \[\Delta \mu \]denote the \[\text{RI}\]of 0th layer and the difference in RI between any two consecutive layers, respectively. The integer \[m=0,1,2,3,...\]denotes the numbers   of the successive layers. A ray of light from the 0th layer enters the 1st  layer at an angle of incidence of \[{{30}^{o}}.\] After undergoing the \[mth\]refraction, the ray emerges parallel to the interface. If \[\mu =1.5\]and \[\Delta \mu =0.015,\]the value of\[m\]is

    A)  20

    B)  30

    C)  40

    D)  50

    Correct Answer: D

    Solution :

     By Snells law, \[\mu \sin \,i=\,\text{constant}\] \[\mu \,\sin i=(\mu -m\Delta \mu )\sin \,r\] Where \[\mu =1.5,\,i={{30}^{o}},\,r={{90}^{o}},\,\,\Delta \mu =0.015\] \[1.5\,\sin \,{{30}^{o}}=(1.5-m\times 0.15)sin{{90}^{o}}\]   \[\frac{3}{2}\times \frac{1}{2}=(1.5-m\times 0.015)\times 1\] \[\frac{3}{4}=\frac{3}{2}-\frac{15m}{1000}\] \[15\,m=\left( \frac{3}{2}-\frac{3}{4} \right)\times 1000\] \[15\,m=\frac{6-3}{4}\times 1000\Rightarrow m=\frac{3}{4}\times \frac{1000}{15}\] \[m=\frac{3000}{60}\Rightarrow m=50\]


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