Practice portal › Ray Optics and Optical Instruments › Reflection by Plane and Spherical Mirrors

An object pin is kept on the axis of a concave mirror, between the pole (P) and the focal point (F). Its image would be ______.

Asked in GSEB Board July 2022 · Mirror formula and magnification

Answer: (4) Virtual, erect and big

Step-by-step solution

Given: a concave mirror with the object between the pole and the focus.

Idea: write f = -f₀ and u = -u₀ with u₀ < f₀, and let the mirror formula say what happens.

1/v = 1/f-1/u = -1/(f₀)+1/(u₀) = (f₀-u₀)/(u₀ f₀), which is positive because u₀ < f₀.

A positive v puts the image behind the mirror, where no light actually reaches, so the image is virtual.

The magnification is m = f/(f-u) = (f₀)/(f₀-u₀), which is positive and greater than 1: erect and enlarged.

This is the shaving-mirror case -- hold your face inside the focus of a concave mirror and you see an upright, magnified image behind the glass.

Why the other options are wrong

More Reflection by Plane and Spherical Mirrors questionsAll Reflection by Plane and Spherical Mirrors questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer