An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is

Question

An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is (A) 1 m (B) 0.5 m (C) 0.24 m (D) 2 m

Tardigrade Admin · Accepted Answer

Here, m = (v/u) = -4 or u = (-v/4) Also, |u|+|v| = 1.5 (v/4) + v = 1.5 or v = 1.2 m and u = (-1.2/4) = -0.3 m ∴ f = (uv/u-v) = (-03 × 1.2/-0.3 - 1.2) = 0.24 m

Q. An object is placed at a distance of $1.5 m$ from a screen and a convex lens is interposed between them. The magnification produced is $4$ . The focal length of the lens is

1 m

0.5 m

0.24 m

2 m

Here, $m = \frac{v}{u} = - 4$ or $u = \frac{- v}{4}$
Also, $∣ u ∣ + ∣ v ∣ = 1.5$
$\frac{v}{4} + v = 1.5$
or $v = 1.2 m$ and
$u = \frac{- 1.2}{4} = - 0.3 m$
$∴ f = \frac{uv}{u - v}$
$= \frac{- 03 \times 1.2}{- 0.3 - 1.2} = 0.24 m$

Q. An object is placed at a distance of 1.5m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is

1 m

0.5 m

0.24 m

2 m

Here, m=uv​=−4 or u=4−v​ Also, ∣u∣+∣v∣=1.5 4v​+v=1.5 or v=1.2m and u=4−1.2​=−0.3m ∴f=u−vuv​ =−0.3−1.2−03×1.2​=0.24m

Q. An object is placed at a distance of $1.5 m$ from a screen and a convex lens is interposed between them. The magnification produced is $4$ . The focal length of the lens is

Here, $m = \frac{v}{u} = - 4$ or $u = \frac{- v}{4}$
Also, $∣ u ∣ + ∣ v ∣ = 1.5$
$\frac{v}{4} + v = 1.5$
or $v = 1.2 m$ and
$u = \frac{- 1.2}{4} = - 0.3 m$
$∴ f = \frac{uv}{u - v}$
$= \frac{- 03 \times 1.2}{- 0.3 - 1.2} = 0.24 m$