Question

A convex lens has a focal length of the 30 cm.An object placed at a distance of 20 cm from the lens what is the nature of image formed by refraction at lens ?​

Answer 1

Given that,

• Focal length (f) = 30cm
• Object distance (u) = -20cm

To find,

• Nature of image

Solution:

By using lens formula,

$\sf \implies \dfrac{1 }{f} = \dfrac{1}{v} - \dfrac{1}{u} \\ \\ \sf \implies \dfrac{1}{30} = \dfrac{1}{v} - ( \dfrac{1}{ - 20} ) \\ \\ \sf \implies \dfrac{1}{v} = \dfrac{1}{30} - \dfrac{1}{20} \\ \\ \sf \implies \frac{1}{v} = \frac{ - 10}{600} \\ \\ \sf \implies \boxed{ \bf \: v = - 60}$

Negative sign of v shows that image is virtual.

And,

$\sf \: m = \dfrac{ h_{2}}{ h_{1} } = \dfrac{v}{u} \\ \sf \: m = \frac{ - 60}{ - 20} \\ \sf \: \boxed{ \bf \: m = 3}$

Here, m is greater than 1, the image is three times larger than the size of object, magnified and erect.

Answer 2

Answer:

50

Explanation: