Ray Optics

JEE Main level, single correct, medium

123 of 295

Question 123 of 295

An object ABAB is placed 15 cm15\text{ cm} on the left of a convex lens PP of focal length 10 cm10\text{ cm}. Another convex lens QQ is now placed 15 cm15\text{ cm} right of lens PP. If the focal length of lens QQ is 15 cm15\text{ cm}, the final image is _______.

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Answer: (B) real, formed at 7.5 cm7.5\text{ cm} right of lens QQ, with a size same as that of ABAB

For lens PP:

1v1−1u1=1f1\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}

1v1−1−15=110\dfrac{1}{v_1} - \dfrac{1}{-15} = \dfrac{1}{10}

1v1=110−115=130⇒v1=30 cm\dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{1}{30} \Rightarrow v_1 = 30\text{ cm}

Magnification m1=v1u1=30−15=−2m_1 = \dfrac{v_1}{u_1} = \dfrac{30}{-15} = -2

The image formed by lens PP acts as a virtual object for lens QQ. Since the distance between the lenses is 15 cm15\text{ cm}, the object distance for lens QQ is:

u2=30−15=+15 cmu_2 = 30 - 15 = +15\text{ cm}

For lens QQ:

1v2−1u2=1f2\dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f_2}

1v2−115=115\dfrac{1}{v_2} - \dfrac{1}{15} = \dfrac{1}{15}

1v2=215⇒v2=7.5 cm\dfrac{1}{v_2} = \dfrac{2}{15} \Rightarrow v_2 = 7.5\text{ cm}

The positive sign indicates that the final image is real and formed 7.5 cm7.5\text{ cm} to the right of lens QQ.

Magnification m2=v2u2=7.515=12m_2 = \dfrac{v_2}{u_2} = \dfrac{7.5}{15} = \dfrac{1}{2}

Total magnification m=m1×m2=−2×12=−1m = m_1 \times m_2 = -2 \times \dfrac{1}{2} = -1

Since ∣m∣=1|m| = 1, the size of the final image is the same as that of the object ABAB.

Answer: real, formed at 7.5 cm7.5\text{ cm} right of lens QQ, with a size same as that of ABAB

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JEE Main 2025 Apr 2 Shift 1 question paper

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