Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.

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Solution

Here, $r = \infty$ , u = -3 cm

$μ_{1} = \frac{3}{2}$ , $μ_{2} = 1$

$v = ?$

$\frac{1}{v} + \frac{3}{2 \times 3} = \frac{μ_{2} - μ_{1}}{α}$

$v = - 2$

So, shift = 3 - 2 = 1 cm from paper.

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Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.

Here, r=∞, u = -3 cm μ1=32, μ2=1 v=? 1v+32×3=μ2−μ1α v=−2 So, shift = 3 - 2 = 1 cm from paper.

Here, $r = \infty$ , u = -3 cm

$μ_{1} = \frac{3}{2}$ , $μ_{2} = 1$

$v = ?$

$\frac{1}{v} + \frac{3}{2 \times 3} = \frac{μ_{2} - μ_{1}}{α}$

$v = - 2$

So, shift = 3 - 2 = 1 cm from paper.