Question

11. The focal length of a converging lens is 30 cm. An object is 90 cm from

the lens. Where will the image be formed and what kind of image is it​

Answer 1

Answer:

We can use Gaussian equation for thin lens with Cartesian sign convention . 1/f = 1/v - 1/ u

Here, object distance, u = -30 cm.

Focal length , f= 50 cm.

Image distance, v = ?

Thus, 1/50= 1/v -(-1/u) OR

1/50 - 1/30 = 1/v OR

v= - 75 cm.

The image is on the side of object and is erect and virtual . The magnification , m is positive. See figure.

Answer 2

Given:-

• Focal length ,f = 30cm

• Object distance ,u = -90cm

To Find:-

• Image distance ,v

Solution:-

By using lens formula.

• 1/v -1/u = 1/f

Substitute the value we get

→ 1/v = 1/f +1/u

→ 1/v = 1/30 + 1/(-90)

→ 1/v = 1/30-1/90

→ 1/v = 3-1/90

→ 1/v = 2/90

→ 1/v = 1/45

→ v = 45cm

Therefore, the image formed 45 cm on the other side of lens . It is inverted and real.