Question

An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is 15 cm?​

Answer 1

Given:-

• Focal length ,f =10cm

• Object Distance ,u = -15 cm

To Find:-

• Image distance & Nature of image

Solution:-

Firstly we calculate the image distance . Using lens formula

❒• 1/v - 1/u = 1/f

where,

v is the Image Position

u is the object distance

f is the focal length

Substitute the value we get

→ 1/v - 1/(-15) = 1/10

→ 1/v +1/15 = 1/10

→ 1/v = 1/10-1/15

→ 1/v = 3-2/30.

→ 1/v = 1/30

→ v = 30 cm

The image distance is 30 cm.

Now, calculating the Magnification

• m = v/u

Substitute the value we get

→ m = -30/15

→ m = -2

Nature of image :- The image formed is Real and inverted .

Answer 2

Answer:

Given :-

• Focal length = 10 cm
• Object distance = -15 cm

To Find :-

Nature of image

Solution :-

By using lens formula

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

$\sf \: \dfrac{1}{v} - \dfrac{1}{ - 15} = \dfrac{1}{10}$

$\sf \: \dfrac{1}{v} + \dfrac{1}{15} = \dfrac{1}{10}$

$\sf \: \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}$

$\sf \: \dfrac{1}{v} = \dfrac{1}{30}$

$\sf \: v = 30 \: cm$

Image distance = 30 cm

Now,

Let's find magnification

$\sf \: m \: = \frac{ - 30}{ - 15}$

m = -2

Hence :-

• Image distance = 30 cm
• Nature :- Real and inverted