Question

The focal length of a convex lens is 30 cm and the size of the real image is one-fourth of the object, then the object distance is

Answer 1

Explanation:

Given that,

Focal length f=30cm

Image distance =v

Object distance =u

By using lens formula

f

1

=

v

1

+

u

1

We are given that,

h

i

=

4

h

0

h

0

h

i

=

4

1

Where,

h

i

= Image height

h

0

= Object height

Now, as

We get

h

0

h

i

=

u

−v

4

1

=

u

−v

u=−4v

v=

4

−u

Now, using equation formula

30

1

=

u

−4

+

u

1

30

1

=

u

−3

u=−90cm

Hence, The object is in front of mirror at a distance 90 cm

Answer 2

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

Object distance = 90cm

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

It is given that :

Focal length of mirror = 30cm.

Object distance = u = ?

Image distance = v = ?

Size of Object =

$h_o$

Size of Image =

$h_i = \frac{1}{4} \times h_o$

Now we know that ratio of height of image to height of object is equal to ratio of image distance to object distance.

$\frac{h_i}{h_o} = \frac{ - v}{u}$

Now for this case ratio of height of image to height of object is :

$\frac{ \frac{1}{4} \times h_o}{h_o}$

$= \frac{1}{4}$

Therefore we can say that :

$\frac{1}{4} = \frac{ - v}{u}$

From this equation we get:

$v = \frac{ - u}{4}$

Now according to mirror formula :

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

In this equation :

• v = Image distance
• u = Object distance
• f = focal length

Now after inputting the known values in this equation we get :

$\frac{1}{30} = \frac{ - 4}{u} + \frac{1}{u}$

$f = \frac{vu}{v + u}$

Therefore :

$30 = \frac{ \frac{ - u}{4} \times u }{ \frac{ - u}{4} + u}$

$30 = \frac{ \frac{ { - u}^{2} }{4} }{ \frac{ - 3u}{4} }$

$30 = \frac{ {u}^{2} }{3u}$

$90u = {u}^{2}$

$u = 90cm$

Therefore the object distance is 90cm.

______________________________________

BY √\________________ SCIVIBHANSHU

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