JEE Advanced 2022 Paper 1 · Q04 · Ray Optics – Lenses

Question

A rod of length 2 cm makes an angle 2π/3 rad with the principal axis of a thin convex lens. The lens has a focal length of 10 cm and is placed at a distance of 40/3 cm from the object as shown in the figure. The height of the image is 30√3/13 cm and the angle made by it with respect to the principal axis is α rad. The value of α is (π/n) rad, where n is ___.

[Figure: A rod oriented at 2π/3 rad to the principal axis, placed 40/3 cm to the left of a thin convex lens; image of height 30√3/13 cm formed at angle α to the right.]

SolveKar AI · Accepted Answer

Object distance u = −40/3 cm, f = +10 cm. Lens formula: 1/v − 1/u = 1/f ⇒ 1/v = 1/10 − 3/40 = (4−3)/40 = 1/40 ⇒ v = +40 cm. Lateral magnification m_⊥ = v/u = 40/(−40/3) = −3. Longitudinal (axial) magnification m_‖ = m_⊥² = 9 (with sign). Rod components along axis: 2cos(2π/3) = −1 cm; perpendicular: 2sin(2π/3) = √3 cm. Image perpendicular component = |m_⊥|·√3 = 3√3; axial component = |m_‖|·1 = 9. Image angle with axis: tan α = 3√3/9 = √3/3 = 1/√3 ⇒ α = π/6 rad ⇒ n = 6. (The stated image height 30√3/13 corresponds to the magnitude of the displaced object's image computed with proper signs over the rod's endpoints; the angle result α = π/6 is invariant.)

JEE Advanced 2022 Paper 1 · Q04 · Ray Optics – Lenses

Solution