Ray optics

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Question

A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the sphere is

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Solution

Using refraction formula $\frac{_{1} μ_{2} - 1}{R} = \frac{_{1} μ_{2}}{v} - \frac{1}{u}$
in given case, medium (1) is glass and (2) is air
$\begin{array}{l} so \frac{_{g} μ_{a} - 1}{R} = \frac{_{g} μ_{a}}{v} - \frac{1}{u} \Rightarrow \frac{\frac{1}{1 .5} - 1}{- 6} = \frac{1}{1 .5 v} - \frac{1}{- 6} \\ \Rightarrow \frac{1 - 1 .5}{- 6} = \frac{1}{v} + \frac{1 .5}{6} \Rightarrow \frac{0 .5}{6} = \frac{1}{v} + \frac{1}{4} \\ \Rightarrow \frac{1}{v} = \frac{1}{12} - \frac{1}{4} = - \frac{2}{12} = - \frac{1}{6} \Rightarrow v = 6 cm \end{array}$

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Ray optics

Remember concepts with our Masterclasses.

A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the sphere is

Using refraction formula 1μ2−1R= 1μ2v−1uin given case, medium (1) is glass and (2) is airso gμa−1R= gμav−1u⇒11.5−1−6=11.5v−1−6⇒1−1.5−6=1v+1.56⇒0.56=1v+14⇒1v=112−14=−212=−16⇒v=6 cm

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