Question

the image of an object formed by a lens is of magnification - 1 . if the distance between the object and its image is 60 cm what is the focal length of the lens ? if the object is moved 20 cm towards the lens where would the image be formed ?

Answer 1

Hello Dear.

Here is your answer---

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Magnification (m) = -1

∴ m = v/u
-1 = v/u
v = -u
v + u = 0 ------------ eq (1)

According to the Question,

∴ v - u = 60 --------------eq (2)

Solving the equations simultaneously,
v = 30 cm
∴ u = v - 60
= 30 - 60
= -30 cm.

Now, Using the lens Formula,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
⇒$\frac{1}{f} = \frac{1}{30} - \frac{1}{-30}$
⇒f = 15 cm.

Thus, the focal length of the lens is 15 cm.

Now, In second case,

Object is shifted 20 cm towards the lens,
Thus, u = -30 + 20
u = -10 cm

Now, Again using the lens Formula,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{15} = \frac{1}{v} - \frac{1}{-10}$
v = - 30 cm.

Thus, the image is formed 30 cm behind the object i.e., on the same side of the lens.

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Hope it helps. :-) ☺☺☺

Answer 2

For lens ,
magnification = image distance/object distance
e.g., m = v/(-u) [ object always lies left side of lens means negative x axis ]
⇒-1 =- v/u
⇒v - u = 0 ⇒v = u

Given, v + u = 60 cm.
⇒v + v = 60
⇒v = 30 and u = 30 cm

Now, use lens formula,
1/v - 1/u = 1/f
v = 30 cm , u = -30 cm
1/30 + 1/30 = 1/f
⇒2/30 = 1/f
⇒f = 15 cm ∴ lens is convex lens .

Now, object is moved 20cm towards the lens
then, new object distance , u' = 30cm - 20cm = 10cm
focal length , f = 15cm
Image distance , v' = ?
Use lens formula,
1/v - 1/u = 1/f
u = -10cm , f = 15cm
⇒1/v' + 1/10 = 1/15
⇒1/v' = 1/15 - 1/10 = (10 - 15)/150 = 1/-30
⇒v' = -30cm

Hence, image distance = 30cm , left side of lens