Question

Find the nature and focal length of a lens which must be placed in contact with a concave lens of focal length 0.25 m in order that the lens combination may produce a real image 5 times the size of the object 0.2 m from the combination.​

Answer 1

Answer:

Convex lens:

Focal length = f = 10 cm. Power of line = P= 10D -

Con cave lens:

Focal length = -25cm.

Power of the lens = -4D

In combination

Object distance = -20cm

Image distance = v = +100cm

$magnification = \frac{height \: of \: image}{height \: of \: object}$

$= \frac{ - 5}{1 } = \frac{r}{u}$

$- 5 (inverted)= \frac{v}{ - 20}$

$- 5 \times - 20 = v$

$100cm = v$

$By \: lens \: Formula,$

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

$\frac{1}{f} + (\frac{ - 1}{25}) = \frac{1}{100} - (\frac{ - 1}{20} )$

$\frac{1}{f} - \frac{1}{25} = \frac{1}{100} + \frac{1}{20}$

$\frac{1}{f} - \frac{1}{25} = \frac{20 + 100}{100 \times 20}$

$\frac{25 - f}{25f} = \frac{120}{2000}$

(zeros will be canceled)

$\frac{25f}{25 - f} = \frac{200}{12}$

$25 \times 12 \times f = 200(25 - f)$

$300f = 5000 - 200f$

$= 500f - 5000$

$f = 10cm$

$power = \frac{1}{focal \: length} = \frac{100}{f}$

$power \: of \: convex \: lens = \frac{100}{10} = 10d$