Question

A equiconvex lens is cut into two halves along (i) XOX and (ii)YOY as shown in the figure. Let f,ff be the focal lengths of the complete lens,of each half in case (i), and of each half in case (ii), respectively 

Choose the correct statement from the following

• A. f'=f, f"=2f
• B. f'=2f, f"=f
• C. f'=f, f"=f
• D. f'=2f, f"=2f

Answer 1

The Correct Option is A

We have, \(\frac{1}{f}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})\)
Here, the lens is equiconvex. Therefore, both sides of the lens will have the same radius of curvature, only in opposite directions. Let it be R. We will use a minus (−)sign to depict the opposite direction.
Therefore, the formula becomes, \(\frac{1}{f}=(\mu-1)(\frac{1}{R}-\frac{1}{-R})\)
Now, the lens is cut along XOX′. Each half of the lens will still act as an equiconvex lens. Hence, we use the same concept used above to get the focal length of each of the pieces.
We have, \(\frac{1}{f'}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})\)
In putting values, \(\frac{1}{f'}=(\mu-1)(\frac{1}{R}-\frac{1}{-R})\)
Therefore, the focal length of each piece in this case, \(f'=f\)
Now, the lens is cut along TOY′. Hence, each of the pieces becomes a plano-convex lens.
Here, the radius of curvature for the plane side will be ∞.
We have, \(\frac{1}{f''}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})\)
In putting values, \(\frac{1}{f''}=(\mu-1)(\frac{1}{R}-\frac{1}{\infty})\)
This gives,\(\frac{1}{f''}=\frac{\mu-1}{R}\)
Therefore, the focal length of each piece in this case, \(f''=2f\)

so,The correct option is(A): f=f, f=2f.