KVPY Sample Paper KVPY Stream-SX Model Paper-25

  • question_answer
    The xz plane separates two media A and B with refractive indices \[{{\mu }_{1}}\] & \[{{\mu }_{2}}\] respectively. A ray of light travels from A to B. Its directions in the two media are given by the unit vectors, \[{{\vec{r}}_{A}}=a\,\hat{i}+b\,\hat{j}\] & \[{{\vec{r}}_{B}}=\alpha \,\hat{i}+\beta \,\hat{j}\] respectively where \[\hat{i}\] & \[\hat{j}\] are unit vectors in the x & y directions. Then:

    A) \[{{\mu }_{1}}a={{\mu }_{2}}\,\alpha \]

    B) \[{{\mu }_{1}}\,\alpha ={{\mu }_{2}}\,a\]

    C) \[{{\mu }_{1}}\,b={{\mu }_{2}}\,\beta \]

    D) \[{{\mu }_{1}}\,\beta ={{\mu }_{2}}\,b\]

    Correct Answer: A

    Solution :

    We have \[\cos \theta =\frac{\vec{A}\cdot \vec{B}}{ab}\] Similarly here
    \[\cos i=\frac{{{{\vec{r}}}_{A}}\cdot \hat{j}}{|{{r}_{A}}|\cdot |1|}=\frac{(a\hat{i}+b\hat{j})\cdot \hat{j}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]    and \[\cos r=\frac{{{{\vec{r}}}_{B}}\cdot \hat{j}}{|{{r}_{B}}|\cdot |1|}\]\[=\frac{(\alpha \hat{i}+\beta \hat{j})\cdot \hat{j}}{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}}=\frac{\beta }{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}}\]
    From shell's law\[{{\mu }_{1}}\sin i={{\mu }_{2}}\sin r\]\[\Rightarrow \]\[{{\mu }_{1}}\sqrt{1-{{\cos }^{2}}i}={{\mu }_{2}}\sqrt{1-{{\cos }^{2}}r}\]\[\Rightarrow \]\[{{\mu }_{1}}\sqrt{1-\frac{{{b}^{2}}}{{{b}^{2}}+{{a}^{2}}}}={{\mu }_{2}}\sqrt{1-\frac{{{\beta }^{2}}}{{{\alpha }^{2}}+{{\beta }^{2}}}}\]
    \[\Rightarrow \]\[{{\mu }_{1}}\frac{a}{\sqrt{{{b}^{2}}+{{a}^{2}}}}={{\mu }_{2}}\frac{\alpha }{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}}\]
    both \[{{\vec{r}}_{A}}\] & \[{{\vec{r}}_{B}}\] are unit vectors so magnitudes of both vectors is 1 i.e.            \[\sqrt{{{b}^{2}}+{{a}^{2}}}=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}=I\] So \[{{\mu }_{1}}a={{\mu }_{2}}\alpha \]

You need to login to perform this action.
You will be redirected in 3 sec spinner