Question 45 of 295

A concave mirror has radius of curvature of $40cm.$ It is at the bottom of a glass that has water filled up to $5cm$ (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance $d$ from the surface of water. The value of $d$ is close to: (Refractive index of water $=1.33)$

A$13.4cm$B$8.8cm$C$6.7cm$D$11.7cm$

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Answer: (B) $8.8cm$

Given
$u=-5 cm$
$f=\dfrac{-R}{2}=-\dfrac{40}{2}=-20 cm$
Now
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}=\dfrac{1}{-20}=\dfrac{1}{\left(-5\right)}$
$∴\text{v}=+\dfrac{20}{3} cm$
Then
Object distance $d=5+\dfrac{20}{3}=\dfrac{35}{3} cm$ (below water surface)
The image after refraction
$d'=d\left(\dfrac{{\mu }_{2}}{{\mu }_{1}}\right)$
$=\dfrac{35}{3}\left(\dfrac{1}{1.33}\right)$
$=8.77$
${d}^{'}≃8.8 cm$