Question

the distance between an object and its real image produced by a converging lens is 0.72m the magnification is 2 what will be the magnification when the object is moved by 0.04m towards the lens​?

Answer 1

The magnification of the object is 4.

Explanation:

Given distance between object and image is 0.72

Hence, |u|+|v| = 0.72 ...........(i)

Also, given magnitude of magnification is 2

|m| = ∣u∣ /∣v∣  =2...........(ii)

Solving equation (i) & (ii), we get

|u| = 0.24, and |v| = 0.48

As image is real, u = -0.24 and v = 0.48

Using lens formula,

1/v - 1/u = 1/f  

​1/0.48 - 1/-0.24 = 1/f

f = 0.16 m

New object distance is: u = -0.24+0.04 = -0.2\ mu=−0.24+0.04=−0.2 m

Using lens formula,

1/v'} - 1/u' = 1/f  

1/v′  − 1 /−0.2  =  1/0.16

u' = 0.8 m

Magnitude of magnification is:

m = ∣u∣ /∣v∣  =  0.2 /0.8 =4

What is magnification?

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