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?​


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Answer 1

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

Answer 2

Given :

Focal length of lens = 10cm

Distance of object = 15cm

Type of lens : concex

To Find :

Distance of image and nature of image.

Solution :

❒ Focal length of convex lens is taken positive and that of concave lens is taken negative.

• Distance of image can be calculated by using lens formula$.$

➙ 1/v - 1/u = 1/f

➙ 1/v - 1/(-15) = 1/10

➙ 1/v = 1/10 - 1/15

➙ 1/v = (3 - 2)/30

➙ v = 30/1

➙ v = 30 cm

In order to find nature of image, we need to calculate magnification first.

➛ m = v/u

➛ m = 30/(-15)

➙ m = -2

❖ Nature of image :

• Real
• Inverted
• Enlarged