A narrow beam of light, parallel to the principal axis, falls on a convex lens of power +2.5 D at a height of 2 cm above the optical centre. The tangent of the angle through which the lens bends the beam is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.05
The power of a lens shows how strongly it bends a beam parallel to the principal axis. A ray at height \(h\) from the optical centre is turned through an angle \(\delta\), where \(\tan\delta=h/f\).
\( \begin{aligned} f &= \frac{1}{P} \\ &= \frac{1}{2.5}\ \mathrm{m} \\ &= 0.4\ \mathrm{m} \end{aligned} \)
\( h=2\ \mathrm{cm}=0.02\ \mathrm{m} \)
\( \begin{aligned} \tan\delta &= \frac{h}{f} \\ &= \frac{0.02}{0.4} \\ &= 0.05 \end{aligned} \)
Both lengths are in metres, so the ratio has no unit. Hence, the tangent of the angle of bending is 0.05.
NCERT (2026-27):
Power of a lens – “… The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light parallel to the principal axis falling at unit distance from the optical centre …”
\( \displaystyle \tan\delta=\frac{h}{f} \)\( ;\ \text{if } h=1,\ \tan\delta=\frac{1}{f} \text{ or } \delta=\frac{1}{f} \text{ for small value of } \delta. \)
“Thus,”
\( \displaystyle P=\frac{1}{f} \)\( \qquad (9.25) \)