VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    A fish, looking up through the water, sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface of water, the radius of the circle in centimetre is

    A)  \[\frac{12\,\times \,3}{\sqrt{5}}\]        

    B)  \[12\,\times \,3\,\times \,\sqrt{5}\]

    C)  \[\frac{12\,\times \,3}{\sqrt{7}}\]        

    D)  \[12\,\times \,3\,\times \,\sqrt{7}\]  

    Correct Answer: C

    Solution :

    \[\tan \,{{i}_{c}}=\frac{r}{h}\] or \[r=h\,\tan {{i}_{c}}\] or  \[r=h\frac{\sin \,{{i}_{c}}}{\cos \,{{i}_{c}}}\]or \[r=h\frac{\sin \,{{i}_{c}}}{\sqrt{1-{{\sin }^{2}}{{i}_{c}}}}\] But \[\sin \,{{i}_{c}}=\frac{1}{\mu }\] \[\therefore \] \[r=h\frac{\frac{1}{\mu }}{\sqrt{1-\frac{1}{{{\mu }^{2}}}}}\] \[=\frac{h}{\sqrt{{{\mu }^{2}}-1}}\] \[=\frac{12}{\sqrt{\frac{16}{9}-1}}=\frac{12\times 3}{\sqrt{7}}cm\]


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