Question

An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens. ​

Answer 1

Solution :

$\underline{\bf{Given\::}}}$

• Distance of Image from lens, (v) = 10 cm
• Height of object, (O) = 50 cm
• Height of Image, (I) = 20 cm

$\underline{\bf{Explanation\::}}}$

As we know that formula of the linear magnification;

$\mapsto\bf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object \:(O)} =\dfrac{Distance\:of\:Image }{Distance\:of\:object} =\dfrac{v}{u} }$

A/q

$\mapsto\sf{\dfrac{I}{O} =\dfrac{v}{u} }\\\\\\\mapsto\sf{\dfrac{20}{50} =\dfrac{10}{u} }\\\\\mapsto\sf{ 20u = 500}\\\\\mapsto\sf{ u = \cancel{500/20}}\\\\\mapsto\bf{u=25\:cm}$

∴ Distance of object from lens, (u) = -25 cm.

Now, using formula of the lens;

$\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u}}}}$

$\longrightarrow\sf{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }\\\\\\\longrightarrow\sf{\dfrac{1}{f} =\dfrac{1}{10} -\dfrac{1}{(-25)} }\\\\\\\longrightarrow\sf{\dfrac{1}{f} =\dfrac{1}{10} + \dfrac{1}{25} }\\\\\\\longrightarrow\sf{\dfrac{1}{f} =\dfrac{5+2}{50} }\\\\\\\longrightarrow\sf{\dfrac{1}{f} =\dfrac{7}{50} }\\\\\longrightarrow\sf{7f=50}\\\\\longrightarrow\sf{f=\cancel{50/7}}\\\\\longrightarrow\bf{f=7.14\:cm}$

Thus;

The focal length of the lens will be 7.14 cm .

Answer 2

Distance of the image from the lens= 10cm

Height of the object(I) = 50 cm

Height of the image(O) =20 cm

We know,

m= Height of the object(I) /Height of the image(O)

=>I/O=V/O

=>20/50=10/U

=>20U=50×10

=>20U=500

=>U=500/20

=>U=25

THEREFORE,

DISTANCE OF THE LENS(U) =-25cm

NOW,

1/F=1/V-1/U

=>1/F=1/10-1/(-25)

=>1/F=1/10+1/25

=>1/F=5+2/50

=>1/F=7/50

=>7F=50

=>F=50/7

=>F=7.14

THEREFORE, THE FOCAL LENGTH=7.14 cm

HOPE IT HELPS YOU....