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Physics Question on Ray optics and optical instruments

An equiconvex lens is cut into two halves along (i) XOX′XOX' and (ii) YOY′YOY' as shown in the figure. Let f,f′,f′′{f, f', f'' }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"=f{f' = f, f" = f}

B

f′=2f,f"=2f{f' = 2f, f" = 2f}

C

f′=f,f"=2f{f' = f, f" = 2f}

D

f′=2f,f"=f{f' =2f, f" = f}

Answer

f′=f,f"=2f{f' = f, f" = 2f}

Explanation

Solution

The focal length of equiconvex lens
1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu -1) \bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)...(i)

1f=(μ−1)(1R−1−R)=2(μ=−1)R \frac{1}{f} = (\mu -1) \bigg(\frac{1}{R}-\frac{1}{-R}\bigg) = \frac{2 (\mu =- 1)}{R}

Case I When lens is cut along XOX′XOX', then each half is again equiconvex with
R1=R,R2=−RR_1 = R, R_2 = - R
∴1f=(μ−1)[1R−1(−R)]\therefore \frac{1}{f} = (\mu -1) \bigg[\frac{1}{R}-\frac{1}{(-R)}\bigg]
=(μ−1)[1R+1R]= (\mu -1) \bigg[\frac{1}{R}+\frac{1}{R}\bigg]
=(μ−1)2R+1f′⇒  f′=f= (\mu -1) \frac{2}{R}+\frac{1}{f'} \Rightarrow \, \, f' = f

Case II When lens is cut along YOY′YOY', then each half becomes piano-convex with
R1=R,R2=8R_1 = R, R_2 = 8
∴1f′′=(μ−1)(1R1−1R2)\therefore \frac{1}{f''} = (\mu -1) \bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)
=(μ−1)(1R+18)= (\mu -1) \bigg(\frac{1}{R}+\frac{1}{8}\bigg)
=(μ−1)R=12f= \frac{(\mu -1)}{R}=\frac{1}{2f}

Hence f′′=2f f'' = 2f