A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam $12$ cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length $20$ cm, and (b) a concave lens of focal length $16$ cm?

Question

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Answer 1

$(a)$ . According to the given case, object is virtual, $u = + 12 c m$
The focal length of the convex lens is $f = + 20 c m$ .

Using lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{1 2} = \frac{1}{2 0}$

$v = 7.5 c m$

$(b)$ . Now the focal length of the concave lens, $f = - 16 c m$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{1 2} = \frac{1}{- 1 6}$

$v = 48 c m$

Answer 2

(a)

. According to the given case, object is virtual,

u = + 12 c m

The focal length of the convex lens is

f = + 20 c m

.

Using lens equation,

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

\frac{1}{v} - \frac{1}{1 2} = \frac{1}{2 0}

v = 7.5 c m

(b)

. Now the focal length of the concave lens,

f = - 16 c m

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

\frac{1}{v} - \frac{1}{1 2} = \frac{1}{- 1 6}

v = 48 c m