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JEE Advanced 2022 · Paper 1, Physics question 4

A rod of length 2 cm makes an angle 2π3\dfrac{2\pi}{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 403\dfrac{40}{3} cm from the object as shown in the figure. The height of the image is 30313\dfrac{30\sqrt3}{13} cm and the angle made by it with respect to the principal axis is α\alpha rad. The value of α\alpha is πn\dfrac\pi n rad, where nn is _____.

A 2 cm rod with its lower end on the principal axis, tilted back over the axis at angle 2π/3 to it. A thin convex lens stands 40/3 cm from the rod's lower end. On the other side, the image is a straight segment with one end on the axis, of height 30√3/13 cm, making angle α with the axis.
Answer and worked solution

Official answer: 5.95

Object ends. The lower end of the rod is on the axis at u=−403u=-\frac{40}3 cm. The rod leans away from the lens at 2π3\frac{2\pi}3 to the axis, so its upper end is

  • 1 cm further from the lens: u=−433u=-\frac{43}3 cm;
  • at height 3\sqrt3 cm.

Images (f=10f=10 cm):

  • lower end: 1v=110−340\frac1v=\frac1{10}-\frac3{40}, so v=40v=40 cm, on the axis;
  • upper end: 1v=110−343\frac1v=\frac1{10}-\frac3{43}, so v=43013v=\frac{430}{13} cm. The magnification is −3013-\frac{30}{13}, giving height 30313\frac{30\sqrt3}{13} cm below the axis, as stated.

Angle. The image ends are 40−43013=901340-\frac{430}{13}=\frac{90}{13} cm apart along the axis:
tan⁡α=303/1390/13=13  ⟹  α=π6.\tan\alpha=\frac{30\sqrt3/13}{90/13}=\frac1{\sqrt3}\implies\alpha=\frac\pi6.

So n=6n=6.

Topics:Thin lenses and lens maker's formula
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