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PhyMedMain 2026 • 5th Apr - Shift 2

Final Image Position and Size for Two Lens Combination

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 ________.

Options

A

virtual, formed at 7.5 cm7.5\text{ cm} right of lens QQ, with a size bigger than that of ABAB

B

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

Correct
C

formed at infinity.

D

real, formed at 7 cm7\text{ cm} right of lens QQ, with a size smaller than that of ABAB

Topics & Concepts

Ray Optics and Optical InstrumentsLenses and Mirrors

Step-by-Step Solution

To determine the characteristics and position of the final image formed by the combination of two lenses, we analyze the refraction of light through each lens sequentially.

Step 1: Refraction through the first convex lens PP

Given:

  • Focal length of lens PP, f1=+10 cmf_1 = +10\text{ cm}
  • Object distance for lens PP, u1=−15 cmu_1 = -15\text{ cm}

Using the thin lens formula: 1v1−1u1=1f1\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}

Substitute the given values: 1v1−1−15=110\frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10} 1v1+115=110\frac{1}{v_1} + \frac{1}{15} = \frac{1}{10} 1v1=110−115=3−230=130\frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} v1=+30 cmv_1 = +30\text{ cm}

The first image I1I_1 is formed at a distance of 30 cm30\text{ cm} to the right of lens PP.

The linear magnification m1m_1 due to lens PP is: m1=v1u1=30−15=−2m_1 = \frac{v_1}{u_1} = \frac{30}{-15} = -2

Step 2: Refraction through the second convex lens QQ

Given:

  • Distance between lens PP and lens QQ, d=15 cmd = 15\text{ cm}
  • Focal length of lens QQ, f2=+15 cmf_2 = +15\text{ cm}

The image I1I_1 acts as a virtual object for lens QQ. Since I1I_1 is formed 30 cm30\text{ cm} to the right of PP and QQ is located 15 cm15\text{ cm} to the right of PP, the position of I1I_1 relative to QQ is: u2=v1−d=30 cm−15 cm=+15 cmu_2 = v_1 - d = 30\text{ cm} - 15\text{ cm} = +15\text{ cm}

Using the thin lens formula for lens QQ: 1v2−1u2=1f2\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}

Substitute the values: 1v2−1+15=115\frac{1}{v_2} - \frac{1}{+15} = \frac{1}{15} 1v2=115+115=215\frac{1}{v_2} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} v2=+152 cm=+7.5 cmv_2 = +\frac{15}{2}\text{ cm} = +7.5\text{ cm}

Since v2v_2 is positive, the final image is real and formed at 7.5 cm7.5\text{ cm} to the right of lens QQ.

Step 3: Total Magnification and Size of the Final Image

The linear magnification m2m_2 due to lens QQ is: m2=v2u2=7.515=+0.5m_2 = \frac{v_2}{u_2} = \frac{7.5}{15} = +0.5

The total magnification mm of the two-lens combination is: m=m1×m2=(−2)×(0.5)=−1m = m_1 \times m_2 = (-2) \times (0.5) = -1

Since ∣m∣=1|m| = 1, the magnitude of the size of the final image is equal to the size of the object ABAB.

Conclusion:

setup leads to a final image that is real, formed at 7.5 cm7.5\text{ cm} right of lens QQ, with a size same as that of ABAB.

Correct Option: B

Final Image Position and Size for Two Lens Combination | Physics PYQ Solution - JEE Challenger