JEE Challenger
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PhyMedAdv 2021 • Paper 1

Total Magnification of Combination of Convex and Concave Lenses

An extended object is positioned at point OO, 10 cm10\text{ cm} in front of a convex lens L1L_1, while a concave lens L2L_2 is located 10 cm10\text{ cm} behind L1L_1, as illustrated in the figure. All curved surfaces of both lenses have radii of curvature equal to 20 cm20\text{ cm}. Both lenses have a refractive index of 1.51.5. Calculate the overall magnification produced by this combination of lenses.

Question Diagram 1

Options

A

0.4

B

0.8

Correct
C

1.3

D

1.6

Topics & Concepts

Ray Optics and Optical InstrumentsLenses and Mirrors

Step-by-Step Solution

To find the total magnification of the combination of lenses, we first determine the focal length of each lens using the Lens Maker's Formula:

1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)

  1. For the convex lens (L1L_1): Here, R1=+20 cmR_1 = +20\text{ cm} and R2=−20 cmR_2 = -20\text{ cm} with μ=1.5\mu = 1.5. 1f1=(1.5−1)(120−1−20)=0.5×220=120 cm−1\frac{1}{f_1} = (1.5 - 1)\left( \frac{1}{20} - \frac{1}{-20} \right) = 0.5 \times \frac{2}{20} = \frac{1}{20}\text{ cm}^{-1} f1=+20 cmf_1 = +20\text{ cm}

  2. For the concave lens (L2L_2): Here, R1=−20 cmR_1 = -20\text{ cm} and R2=+20 cmR_2 = +20\text{ cm} with μ=1.5\mu = 1.5. 1f2=(1.5−1)(1−20−120)=0.5×(−220)=−120 cm−1\frac{1}{f_2} = (1.5 - 1)\left( \frac{1}{-20} - \frac{1}{20} \right) = 0.5 \times \left(-\frac{2}{20}\right) = -\frac{1}{20}\text{ cm}^{-1} f2=−20 cmf_2 = -20\text{ cm}

  1. Refraction at Lens L1L_1: The object is placed at u1=−10 cmu_1 = -10\text{ cm} in front of L1L_1. Using the thin lens formula: 1v1−1u1=1f1\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} 1v1−1−10=120\frac{1}{v_1} - \frac{1}{-10} = \frac{1}{20} 1v1=120−110=−120 cm−1\frac{1}{v_1} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20}\text{ cm}^{-1} v1=−20 cmv_1 = -20\text{ cm}

    The magnification produced by L1L_1 is: m1=v1u1=−20−10=+2m_1 = \frac{v_1}{u_1} = \frac{-20}{-10} = +2

  1. Refraction at Lens L2L_2: The image I1I_1 formed by L1L_1 acts as a virtual object for L2L_2. Since I1I_1 is formed 20 cm20\text{ cm} to the left of L1L_1 and L2L_2 is located 10 cm10\text{ cm} behind L1L_1, the position of this image relative to L2L_2 is: u2=−(20+10)=−30 cmu_2 = -(20 + 10) = -30\text{ cm}

    Using the thin lens formula for L2L_2: 1v2−1u2=1f2\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} 1v2−1−30=1−20\frac{1}{v_2} - \frac{1}{-30} = \frac{1}{-20} 1v2=−120−130=−560=−112 cm−1\frac{1}{v_2} = -\frac{1}{20} - \frac{1}{30} = -\frac{5}{60} = -\frac{1}{12}\text{ cm}^{-1} v2=−12 cmv_2 = -12\text{ cm}

    The magnification produced by L2L_2 is: m2=v2u2=−12−30=+0.4m_2 = \frac{v_2}{u_2} = \frac{-12}{-30} = +0.4

  1. Total Magnification of the Combination: The total magnification mm of the system is the product of the individual magnifications: m=m1×m2=2×0.4=0.8m = m_1 \times m_2 = 2 \times 0.4 = 0.8

Thus, the overall magnification produced by the combination of lenses is 0.8.

Correct Answer: (B) 0.8

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