Question

You have a convex lens of focal length 12cm You place a candle flame on its principal axis at a distance of 60cm from the lens on its one side and place a screen on its other side Write two characteristics of the image formed on the screen If you now shift the candle flame towards the lens so thats its difference from the lens becomes 15cm then state the changes that are observed in these two characteristics of the image formed


plz answer if u know it otherwise i will report​

Answer 1

Answer:

(a) The type of the lens is convex lens and its focal length is 20 cm.

(b) The nature of the image is virtual, erect and enlarged.

Given:

Object distance = u = - 40 cm

Image distance = v = 40 cm

To find:

Type of the lens = ?

Focal length = ?

Nature of the image after the shift = ?

Formula used:

Lens formula:

$</p><p>\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}} </p><p>$

Magnification formula:

$\bold{m=\frac{v}{u}}</p><p>$

Solution:

(a) Focal length is:

$\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}} </p><p>$

$\bold{\frac{1}{f}=\frac{1}{40}-\frac{1}{(-40)}} </p><p>$

$\bold{\frac{1}{f}=\frac{1}{40}+\frac{1}{40}} </p><p>$

$</p><p>\bold{\frac{1}{f}=\frac{2}{40}=\frac{1}{20}}$

$\bold{\therefore f= 20 \ cm}$

Since, the focal length has positive sign, it it a convex lens.

(b) Nature of the image formed when candle flame is shifted 25 cm:

Distance of the object =

$\bold{-(40-25)=-15 \ cm}$

Now, image distance is:

$\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}} </p><p>$

$\bold{\frac{1}{20}=\frac{1}{v}-\frac{1}{(-15)}}$

$\bold{\frac{1}{20}=\frac{1}{v}+\frac{1}{(15)}} </p><p>$

$\bold{\frac{1}{20}-\frac{1}{15}=\frac{1}{v}} </p><p>$

$\bold{\frac{1}{v}=\frac{3-4}{60}=-\frac{1}{60}} </p><p>$

$\bold{\therefore v=-60 \ cm}</p><p>$

Now, we need to find the magnification:

$\bold{m=\frac{v}{u}}$

$\bold{m=\frac{-60}{-15}=4}$

Since, the magnification is positive, then the image is virtual, erect and enlarged.

Answer 2

This answer is search by google.....