Question

An object is placed at a distance of 10 cm from a convex lens of focal length 15 cm. find:

a) Image distance

b) Nature of image

c) Magnification​

Answer 1

Explanation:

image distance . i think sooo

Answer 2

Given:

The distance of the object(u) = 10 cm.

Focal length (f) = 15 cm.

To find:

• The position of the image.

• The nature of the image.
• The magnificent of the image.

Solution:

Let, the distance of the image from the lens be, 'v'.

The magnificent of the image be, 'm'.

The size of the image be, 'I'.

f= 15 cm

u = 10 cm

From the lens formula we get that,

$\dfrac{1}{f} = (\dfrac{1}{u} - \dfrac{1}{v})$

$\\ or, \dfrac{1}{15} = (\dfrac{1}{10}-\dfrac{1}{v})$

$or, (\dfrac{1}{10} - \dfrac{1}{v}) = \dfrac{1}{15}$

$or, (-\dfrac{1}{v})= ( \dfrac{1}{15} - \dfrac{1}{10})$

$or, (-\dfrac{1}{v}) = (\dfrac{2-3}{30})$

$or, (-\dfrac{1}{v})= \dfrac{-1}{15}$

$or, \dfrac{1}{v} = \dfrac{1}{15}$

or, v = 15 cm.

∴ The image has formed at a distance of 15 cm in front of the lens.

From the formula we get that,

$m = \dfrac{-v}{u}$

$or, m = \dfrac{-(15)}{5}$

or, m = -3 [Magnificent is a ratio so, it has no unit.]

∴The magnificent  of the image is  -3.

Since, the magnificent of the image is negative so, the image is real, inverted,  and smaller in size.

Answer:

a)The image has formed 15 cm in front of the mirror.              

b)The image is real, inverted and smaller in size.

c)The magnificent of the image is -3.