Q158 / 170 · Manipal (MET) 2016 · Physics

A thin walled glass sphere of radius 2.5 cm is filled with water. An object $(O)$ is placed at 7.5 cm from the surface of the sphere. Neglecting the effect of glass wall, at what distance the image ( $I$ ) of the object, measured from the centre of sphere is formed? Figure

  • A
    20 cm
  • B
    15 cm
  • Correct answer
    C
    10 cm
  • D
    7.5 cm

Solution

For refraction at first surface, $u=-7.5 \mathrm{~cm}, R_1=2.5 \mathrm{~cm}$ $\mu_1=1, \mu_2=\frac{4}{3}$ $\therefore \quad \frac{\mu_2}{v^{\prime}}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R_1}$ $\frac{4 / 3}{v^{\prime}}-\frac{1}{(-7.5)}=\frac{4 / 3-1}{25}$ $\Rightarrow \quad \frac{4}{3 v^{\prime}}=\frac{1}{7.5}-\frac{1}{7.5}=0$ $\Rightarrow \quad v^{\prime}=\infty$ It means the ray is parallel within the sphere. For refraction at second surface, $u=-\infty, \mu_1=\frac{4}{3}$ $\mu_2=1, R_2=-2.5 \mathrm{~cm}$ $\therefore \quad \frac{\mu_2}{v^{\prime}}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R_2}$ gives $\Rightarrow \quad \frac{1}{v^{\prime}}-\frac{4 / 3}{(-\infty)}=\frac{1-(4 / 3)}{(-2.5)}$ $\Rightarrow \quad v=+7.5 \mathrm{~cm}$ i.e. Image is formed at distance 7.5 cm to the right of $P_2$. Hence, distance of final image $P$ from centre $C=7.5+2.5=10 \mathrm{~cm}$.

Hence, the correct answer is option: 'C'.

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