Q142 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Physics
An object is fixed at the bottom of a vessel and water is filled in the vessel upto a height of $10 \mathrm{~cm}$. A plane mirror is placed at a height of $7 \mathrm{~cm}$ from the surface of water in such a way that its reflecting surface faces the water. The distance of the image from the mirror is (Refractive index of water, $n=1.33$ )
- A7.5 cm
- B7 cm
- Correct answerC14.5 cm
- D21.8 cm
Solution
According to the question,
the apparent distance of object from the mirror, distance of image from the mirror, $d$ $$ =7 \mathrm{~cm}+\text { apparent depth } $$ $$ \begin{aligned} \because \text { Apparent depth } & =\frac{\text { Real depth }}{\mu} \\ & =\frac{10}{1.33} \quad(\because \mu=1.33 \text {, given }) \end{aligned} $$ Hence, $d=7 \mathrm{~cm}+\frac{10}{1.33}=14.5 \mathrm{~cm}$
the apparent distance of object from the mirror, distance of image from the mirror, $d$ $$ =7 \mathrm{~cm}+\text { apparent depth } $$ $$ \begin{aligned} \because \text { Apparent depth } & =\frac{\text { Real depth }}{\mu} \\ & =\frac{10}{1.33} \quad(\because \mu=1.33 \text {, given }) \end{aligned} $$ Hence, $d=7 \mathrm{~cm}+\frac{10}{1.33}=14.5 \mathrm{~cm}$
Hence, the correct answer is option: 'C'.