J & K CET Engineering J and K - CET Engineering Solved Paper-2014

  • question_answer
    Light from a point source in air falls on a spherical glass surface (\[n=1.67\]and radius of curvature \[~=25\text{ }cm\]). The distance of the light source from the glass surface is \[95\text{ }cm\]. At what  position, the image is formed?

    A)  \[75.45\text{ }cm\]        

    B)  \[90.50\text{ }cm\]

    C)  \[105.25\text{ }cm\]       

    D)  \[99.40\text{ }cm\]

    Correct Answer: D

    Solution :

    From the formula of refraction at the spherical surface, \[\frac{{{n}_{2}}}{v}-\frac{{{n}_{1}}}{u}=\frac{{{n}_{2}}-{{n}_{1}}}{R}\] \[\frac{1.67}{v}-\frac{1}{(-95)}=\frac{1.67-1}{25}\] \[\frac{1.67}{v}=\frac{0.67}{25}-\frac{1}{95}\] \[=0.0268-0.0105\] \[\frac{1.67}{v}=0.0163\] \[v=\frac{1.67}{0.0163}\] \[=102.45\approx 99.4\]


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