Question

An object is placed 4cm in front of a convex lens produces a virtual and erect image 12cm from the lens on the same side of the object. What is the magnification of the image . What os the focal length of the odject?​

Answer 1

Answer:

Focal length of lens is 6 cm

Magnification of image = 3

Explanation:

u = - 4 cm

v = - 12 cm

Lens formulae : 1 /v - 1 /u = 1 /f

1 /f = - 1 /12 + 1 /4

     = 2 /12

f = 6 cm

Magnification : m = v /u

m = - 12 /- 4

   = 3

Answer 2

Correct Question :

Q. An object placed 4cm in front of a Converging lens produce a real image 12cm from the lens.

• a) What is the magnification of the image.
• b) What is the focal length of the lens.

Solution :

Let's

Sign Conversation.

• V = +12 Cm.
• U = -4 Cm.
• f = ?

$\rule{90}1$

a) We have, formula

$\tt \dagger \: \: \: m = \frac{v}{u} </p><p> \frac{h_i}{h_o}$

• V distance of image.
• U distance of object.
• hi height of image.
• ho height of object.

$\tt \longmapsto m = \frac{12}{ - 4}$

$\tt \longmapsto m = - 3.$

It means,

$\tt \longmapsto\frac{ h_i}{ h_o }= m$

$\tt \longmapsto - 3 = \frac{ h_i}{ h_o }$

$\tt \longmapsto h_i = 3h_o$

$\rule{90}1$

b) We have Lens Formula.

$\tt \dagger \: \: \: \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\tt \longmapsto \frac{1}{f} = \frac{1}{12} - \frac{1}{ - 4}$

$\tt \longmapsto \frac{1}{f} = \frac{1}{12} + \frac{1}{4}$

$\tt \longmapsto \frac{1}{f} = \frac{1 + 3}{12}$

$\tt\longmapsto \frac{1}{f} = \frac{4}{12}$

$\tt \longmapsto \frac{1}{f} = \frac{1}{3}$

$\tt\longmapsto {f} = + 3 \: cm.$

$\rule{200}3$

Note : Diagram provide above.