A cylindrical glass rod has its two coaxial ends of spherical from bulging outward. The front and has a radius of curvature $5 c m$ and the back end which is silvered has a radius of curvature $8 c m$ . The thickness of the rod along the axis is $10 c m$ . Calculate the position of the image of a point at the axis $50 c m$ from front face $(_{a} n_{g} = 1.5)$

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Answer 1

Refraction at 1st surface :
$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

Substituting $u = - 50 c m, R = 5 c m, μ_{2} = 1.5, m u_{1} = 1$

$\frac{1 . 5}{v} - \frac{1}{- 5 0} = \frac{1 . 5 - 1}{5}$

$\Rightarrow v = 18.75 c m$

This image acts as virtual object for second surface,
Final image distance from second surface, $v^{'}$
$∴ \frac{1}{v ^{'}} - \frac{1 . 5}{( 1 8 . 7 5 - 1 0 )} = \frac{1 - 1 . 5}{- 8}$

$\Rightarrow v^{'} = 4.274 c m$

$∴$ Distance of final image from the first surface $= 10 + 4.274 = 14.274 c m$

Answer 2

Refraction at 1st surface :

\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}

Substituting

u = - 50 c m, R = 5 c m, μ_{2} = 1.5, m u_{1} = 1

\frac{1 . 5}{v} - \frac{1}{- 5 0} = \frac{1 . 5 - 1}{5}

\Rightarrow v = 18.75 c m

This image acts as virtual object for second surface,
Final image distance from second surface,

v^{'}

∴ \frac{1}{v ^{'}} - \frac{1 . 5}{( 1 8 . 7 5 - 1 0 )} = \frac{1 - 1 . 5}{- 8}

\Rightarrow v^{'} = 4.274 c m

∴

Distance of final image from the first surface

= 10 + 4.274 = 14.274 c m