CBSE Class 12 · 2015 · Delhi · Geometrical OpticsShort Answer

Use the mirror equation to show that an object placed between

f

and

2 f

of a concave mirror produces a real image beyond

2 f

Solution

Mirror equation (Cartesian sign convention, concave mirror

f < 0

\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u}

Given

f < ∣ u ∣ < 2 f

with

u, f < 0

. At

u = - 2 f

\frac{1}{v} = \frac{1}{- f} - \frac{1}{- 2 f} = - \frac{1}{2 f} \Rightarrow v = - 2 f

. At

u = - f

\frac{1}{v} = \frac{1}{- f} - \frac{1}{- f} = 0 \Rightarrow v \to - \infty

.
So as

∣ u ∣

decreases from

2 f

f

∣ v ∣

increases from

2 f

\infty

, i.e.

2 f < ∣ v ∣ < \infty

The image is real (same side as object) and, since

m = - v / u < 0

, inverted — formed \emph{beyond}

2 f

CBSE Class 12 Board · 2015 · Delhi · Subjective