Question

People commonly use contact lenses to correct their vision. A patient has a near point of 150 cm and, to correct this, wears contact lenses with a focal length of 50 cm Part A What is the refractive power of her contact lenses? Part B How close can she hold a newspaper and still read it while wearing her contact lenses? Part C The indices of refraction for her contact lens, cornea, and the fluid behind her cornea are 1.6, 1.4, and 1.3, respectively. Light is incident from air onto her contact lens at an angle of 25 ∘ from the normal of the surface. At what angle is the light traveling in the fluid behind her cornea?

Answer 1

(A)

The refractive power is,

$$ \begin{aligned} P &=\frac{1}{f} \\ &=\frac{1}{50 \times 10^{-2} \mathrm{~m}} \\ &=2 \text { diaptar } \end{aligned} $$

(B)

The required distance is,

According to lens formula,

$$ \begin{aligned} &\frac{1}{f}=\frac{1}{-150}+\frac{1}{d} \\ &\frac{1}{d}=\frac{1}{50 \mathrm{~cm}}+\frac{1}{150 \mathrm{~cm}} \\ &d=37.5 \mathrm{~cm} \end{aligned} $$

(C)

According to Snell's law,

$$ \begin{aligned} 1 \sin \left(25^{\circ}\right) &=1.6 \sin \theta \\ \theta &=15.3^{\circ} \end{aligned} $$

For next medium

$$ \begin{aligned} 1.6 \sin \left(15.3^{\circ}\right) &=1.4 \sin \theta^{1} \\ \theta^{1} &=17.5^{\circ} \end{aligned} $$

For third interface,

$$ \begin{aligned} 1.4 \sin \left(25^{\circ}\right) &=1.3 \sin \theta^{11} \\ \theta^{11} &=18.9^{\circ} \end{aligned} $$