Ray Optics

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Question 214 of 295

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :

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Answer: (A) -2

The biconvex lens is a combination of two thin plano-convex lenses in contact.
For the first plano-convex lens: μ1=1.5\mu_1 = 1.5, R1=15R_1 = 15 cm, R2=∞R_2 = \infty.
Using lens maker's formula:
1f1=(μ1−1)(1R1−1R2)\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
=(1.5−1)(115−1∞)= (1.5 - 1) \left( \frac{1}{15} - \frac{1}{\infty} \right)
=0.5×115=130= 0.5 \times \frac{1}{15} = \frac{1}{30} cm−1^{-1}.
For the second plano-convex lens:
μ2=1.2\mu_2 = 1.2, R1=∞R_1 = \infty, R2=−12R_2 = -12 cm.
Using lens maker's formula:
1f2=(μ2−1)(1R1−1R2)\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
=(1.2−1)(1∞−1−12)= (1.2 - 1) \left( \frac{1}{\infty} - \frac{1}{-12} \right)
=0.2×112=160= 0.2 \times \frac{1}{12} = \frac{1}{60} cm−1^{-1}.
The equivalent focal length FF of the combination is given by:
1F=1f1+1f2=130+160\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{60}
=2+160=360=120= \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20} cm−1^{-1}.
Thus, F=20F = 20 cm.
Given object distance u=−30u = -30 cm. Using the lens formula 1v−1u=1F\frac{1}{v} - \frac{1}{u} = \frac{1}{F}:
1v−1−30=120\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}
⇒1v=120−130=3−260=160\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}.
So, v=60v = 60 cm.
Magnification m=vu=60−30=−2m = \frac{v}{u} = \frac{60}{-30} = -2.

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