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 :

• Object distance from the convex lens is 15 cm
• Focal length of lens is 10 cm

To Find :

• Nature of image
• Image distance from the lens

Theory :

${\red{\boxed{\large{\bold{Lens\:Formula}}}}}$

$\rm\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Solution :

We have

$\sf\:u=-15cm(by\:sign\:convention)$

$\sf\:f=10cm(by\:sign\:convention)$

and $\sf\:v=?$

Now By using Lens Formula

$\sf\:\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the given values

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

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

$\sf\implies\:\dfrac{1}{v}=\dfrac{3-2}{30}$

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

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

Therefore, Image distance is 30 cm from the convex lens.

In this case , image will formed beyond 2F.Therefore , the nature of image is real & inverted .

Answer 2

GiveN :-

• Distance of object from convex lens = - 15 cm

• Focal length of the lens = 10 cm

To FinD :-

• Distance of the image from the lens

• Nature of the image

SolutioN :-

By using lens formula

$\large:\implies \underline { \boxed { \green{\bf \frac{1}{f} = \frac{1}{v} - \frac{1}{u}}}} \\ \\ : \implies \sf \frac{1}{10} = \frac{1}{v} - \frac{1}{ - 15} \\ \\: \implies \sf \frac{1}{v} = \frac{1}{10} - \frac{1}{15} \\ \\: \implies \sf \frac{1}{v} = \frac{3 - 2}{30} \\ \\: \implies \sf \frac{1}{v} = \frac{1}{30} \\ \\: \implies \boxed{ \sf v = 30 \: cm}$

As image will formed beyond 2F. Therefore , the nature of image is real & inverted