Question

5. An object is placed at a distance of 50cm from a concave lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image. Calculate focal length of the lens and magnification produced.

Answer 1

Check this attachment :-)

Answer 2

According to the Question

u = -50cm

v = -10cm (Image is Virtual)

f = To find

Using Lens formula

$\bf\huge\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\bf\huge\frac{1}{f} = \frac{1}{-10} - \frac{1}{-50}$

$\bf\huge\frac{1}{f} = \frac{-5 + 1}{50}$

$\bf\huge\frac{1}{f} = \frac{-4}{50}$

$\bf\huge f = \frac{-50}{4}$

f = -12.5 cm

Negative sign show that lens is concave

Focal length = -12.5 cm