An object placed in front of a concave mirror at a distance of x cm from the pole gives a 3 times magnified real image. If it is moved to a distance of (x+5) cm, the magnification of the image becomes 2. The focal length of the mirror is

Question

An object placed in front of a concave mirror at a distance of x cm from the pole gives a 3 times magnified real image. If it is moved to a distance of (x+5) cm, the magnification of the image becomes 2. The focal length of the mirror is (A) 15 cm (B) 20 cm (C) 25 cm (D) 30 cm

Tardigrade Admin · Accepted Answer

Given v1=3 x and v2=2(x+5) So here, Ist case (1/-x)+(1/-3 x)=(1/f) ...(i) (1/-(x+5))+(1/-2(x+5))=(1/f) ...(ii) On solving Eqs. (i) and (ii) we get f=30 cm

Q. An object placed in front of a concave mirror at a distance of $x c m$ from the pole gives a $3$ times magnified real image. If it is moved to a distance of $(x + 5) c m$ , the magnification of the image becomes $2$ . The focal length of the mirror is

15 cm

20 cm

25 cm

30 cm

Given $v_{1} = 3 x$ and $v_{2} = 2 (x + 5)$
So here, Ist case
$\frac{1}{- x} + \frac{1}{- 3 x} = \frac{1}{f}$ ...(i)
$\frac{1}{- ( x + 5 )} + \frac{1}{- 2 ( x + 5 )} = \frac{1}{f}$ ...(ii)
On solving Eqs. (i) and (ii) we get $f = 30 c m$

Q. An object placed in front of a concave mirror at a distance of xcm from the pole gives a 3 times magnified real image. If it is moved to a distance of (x+5)cm, the magnification of the image becomes 2. The focal length of the mirror is