Question

For an object placed at a distance $24\, m$ from a lens, a sharp focused image is observed on a screen placed at a distance $12\, cm$ from the lens A glass plate of refractive index $15$ and thickness $1 \,cm$ is introduced between lens and screen such that the glass plate plane faces parallel to the screen By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

• A. 0.8 m
• B. 3.2 m
• C. 1.2 m
• D. 5.6 m

Answer 1

Answer: (B)

3.2 m

Answer 2

Applying lens formula
0.121​+2.41​=f1​⇒f1​=24210​
Upon putting the glass slab, shift of image is
Δx=t(1−μ1​)=31​cm
Now v=12−31​=335​cm
Again apply lens formula
0.121​+u1​=f1​=24210​
Solving u=−5.6m
Thus shift of object is
5.6−2.4=3.2m