Question

When an object of height is 3 cm is placed at a distance of 30 cm from a lens, a real image is formed at a distance of 60 cm. Find out the height of the image

Answer 1

AnsWer :

Height of the image be 6 cm.

Concepts Required :

What is means of Real images ?

• Image formed ( Real and Inverted )
• Inverted ( height of object opposite to the height of image. )
• Example - Our eyes.

Solution :

We have formula,

$\tt\dagger \: \: \: \: \: m = \frac{ - v}{u} = \frac{h_i}{h_o}$

Where,

• v image distance.
• u Object distance.
• hi height of image.
• ho height of object.

Now, Putting given value.

$\tt \longmapsto m = \frac{ - {}\cancel{6}\cancel0}{\cancel{3}\cancel0} = \frac{h_i}{3}$

$\tt \longmapsto m = - 2 = \frac{h_i}{3}$

$\tt\longmapsto - 6 = h_i$

$\tt\longmapsto h_i = - 6 \: cm.$

Therefore, the height of image be 6 cm.

$\rule{80}1$

Note : More information provide in attachment.

Some information :

• 6 cm negative sign show that height of image be opposite sides of height of object.
• Mirror Formula : 1/f = 1/v + 1/u.
• Lens Formula : 1/f = 1/v - 1/u.
• Magnified : m = -v/u = hi/ho.

Answer 2

Answer:

Explanation: ho =3cm, u=-30cm, v=60cm

m=-v/u

m=hi/ho

so,  

    -v/u=hi/ho

    -60/-30=hi/3cm

if we cross multiply we get

     hi=-60*3/-30

        =-180/-30

        =180/30=6cm

hi=6cm