JEE Advanced 2015 · Paper 1 · Physics Q8

← JEE Advanced 2015 · Paper 1

PhysicsInteger+4Question 8 of 60

Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification $M_1$ . When the set-up is kept in a medium of refractive index 7/6, the magnification becomes $M_2$ . The magnitude $\left|\dfrac{M_2}{M_1}\right|$ is

A concave mirror on the left facing right, an object 15 cm in front of it on the axis, and a convex lens 50 cm from the mirror on the right.

Your answer

Answer and worked solution

Official answer: 7

Mirror. $u=-15$ cm and $f=-10$ cm give $v=-30$ cm, with $m_1=-2$ . The mirror is not affected by the medium.

Lens in air. The mirror's image is $50-30=20$ cm from the lens. With $f=10$ cm the image is at 20 cm and $m=-1$ . So $M_1=(-2)(-1)=2$ , erect.

Lens in the medium (n = 7/6).
$\frac{f'}{f}=\frac{1.5-1}{\frac{1.5}{7/6}-1}=\frac{0.5}{2/7}=\frac74\implies f'=17.5\text{ cm}.$
Object at 20 cm: $\frac1v=\frac1{17.5}-\frac1{20}$ , so $v=140$ cm and $m=-7$ . Then $M_2=(-2)(-7)=14$ .

Ratio. $|M_2/M_1|=7$ .

Answer: 7

Topics:Combination of lenses and mirrors Thin lenses and lens maker's formula Spherical mirrors

← Previous question Next question →

JEE Advanced 2015 · Paper 1, Physics question 8