Question

A square ABCD of side 1mm is kept at distance 15cm infront of the concave mirror . The focal length of the concave mirror is 10cm . The length of the perimeter of its image will be (1)8mm. (2) 2mm (3)12mm (4)6 mm

Answer 1

Answer on Question #74433, Physics / Optics

Question. A square $ABCD$ of side $1 \, \text{mm}$ is kept at distance $15 \, \text{cm}$ in front of the concave mirror. The focal length of the concave mirror is $10 \, \text{cm}$. The length of the perimeter of its image will be

(1) $8\,\text{mm}$;

(2) $2\,\text{mm}$;

(3) $12\,\text{mm}$;

(4) $6\,\text{mm}$.

Given. $h_o = 1\,\text{mm}$; $d_0 = 15\,\text{cm}$; $f = 10\,\text{cm}$

Find. $p - ?$

Solution.

For the concave mirror

$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \rightarrow d_i = \frac{1}{\frac{1}{f} - \frac{1}{d_o}} = \frac{1}{\frac{1}{10} - \frac{1}{15}} = 30\,\text{cm}$$

So,

$$\frac{h_i}{h_o} = -\frac{d_i}{d_o} \rightarrow h_i = -h_o \frac{d_i}{d_o} = -1 \cdot \frac{30}{15} = -2\,\text{mm}$$

Finally

$$p = 4 \cdot h_i = 4 \cdot 2 = 8\,\text{mm}$$

Answer. $p = 8\,\text{mm}$.

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