Question

A lens of focal length 12 cm produces a virtual image,
whose linear dimensions are 3 that of the object. The
distance of object and the image from the lens are
respectively
​

Answer 1

Answer:

U = −16 cm

V = 48 cm

Explanation:

The formula for focal length is;

1/v – 1/u = f

( u−v ) / uv = 12….. let it be equation 1

We are given that magnification is three times

M = −v / u = 3

V =−3u

Using this in equation (1)

(u+3u ) / −u×3u = 12

Solving the above we get

U = −16 cm

And

V = 48 cm

Answer 2

The  distance of object and the image from the lens are  8 cm and -24 cm respectively.

Step-by-step explanation:

The lens formula is given as:

1/f = 1/u + 1/v

Where,

f = Focal length

u = Object distance

v = Image distance

On substituting value of focal length, we get,

1/12 = 1/u + 1/v

1/12 = (v + u)/vu

uv = 12(v + u)

12v + 12u = uv → (equation 1)

From question, magnification is 3 times of the object.

Now, m = -v/u = 3

⇒ v = -3u → (equation 2)

Now, the equation (1) becomes,

12(-3u) + 12u = u(-3u)

-36u + 12u = -3u²

-24u = -3u²

24u = 3u²

8u = u²

u² - 8u = 0

u(u - 8) = 0

∴ u = 8 cm

On substituting value of 'u' in equation (2), we get,

v = -3(8)

∴ v = - 24 cm