Question

 A 5 cm tall object is placed perpendicular to principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. The nature, position and size of the image is​

Answer 1

Answer:

Explanation:

The lens formula is given by $\frac{1}{v} - \frac{1}{u} =\frac{1}{f}$

Given: -

f=10cm

u=15 cm

For a convex lens,

u= negative

v=positive

f = positive

so, modifying the lens formula we have

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

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

Subsituting the value,

$\frac{1}{v} + \frac{1}{15}= \frac{1}{10}$

$\frac{1}{v}= \frac{1}{10} - \frac{1}{15}\\\frac{1}{v} = \frac{3-2}{30}\\\frac{1}{v}=\frac{1}{30}$

v= 30 cm

Nature of the image is real and inverted.

Position of the image is 30 cm on the other side of the lens as the object

For the size of the image,

$\frac{v}{u}=\frac{I}{O}\\ \\\frac{30}{15}= \frac{I}{5}\\\\2= \frac{I}{5}\\ \\10= I$

The image size is 10 cm