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Q.4. A square ABCD of side 1 mm is kept at distance 15 cm infront of the concave mirror as shown in the figure. The focal length of the mirror is 10 cm. The length of the perimeter of its image will be :



(A) 8 mm

(B) 2 mm

(C) 12 mm

(D) 6 mm

Answer 1

Answer:

Perimeter of the image is given as 8mm

Explanation:

As we know that by mirror formula we have

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

so here we will have

$u = 15 cm$

$f = 10 cm$

$\frac{1}{v} + \frac{1}{15} = \frac{1}{10}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{15}$

$\frac{1}{v} = \frac{1}{30}$

$v = 30 cm$

so magnification is given as

$M = \frac{v}{u}$

$M = \frac{30}{15} = 2$

now we have

$A_{image} = m^2A_{object}$

$A_{image} =2^2(1mm)^2$

so length of side of image is

$L = 2 mm$

so total perimeter is given as

$P_{image} = 4(2mm)$

$P_{image} = 8mm$

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Topic : Concave mirror

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