Question

1. A nearsighted person has a far point of 21 cm. To correct the distant vision using a lens 2 cm from the eye. What power lens should be used?

1. 4.8 D

2. -4.8 D

3. 5.3 D

4. -5.3 D

2. The speed of light in a plastic material is 1.5*10^8 m/s. The index of refraction of this material is

A. 2

B.1.54

C. 1.33

D. 0.81

3 if light of wavelength 600 nm in air enters a medium with index of refraction. the wave length in this new medium is

A.Unchanged

B.900 nm

C.400nm

D.601.5 nm

Answer 1

(1) Using a lens formula, we have

1 / d₀ + 1 / d_i = 1 / f

1 / f = 1 / () - 1 / (D_F - D_L)

1 / f = 1 / () - 1 / [(21 - 2) cm]

f = - 19 cm

The refractive power of a lens should be given as :

P (in dioters) = 1 / f (in meter)

P = 1 / (-0.19 m)

P = - 5.26 D

option-4 : it is correct.

(2) The index of refraction of this material will be given as :

using a formula, we have

n = c / v = (3 x 10⁸ m/s) / (1.5 x 10⁸ m/s)

n = 2

(3) If light of wavelength 600 nm in air which enters a medium with index of refraction, then the wavelength in this new medium will be given as :

we know that, n = c / v = ₁ / ₂

where, n = refractive index of air = 1

₂ = (600 nm) / 1

₂ = 600 nm

(Unchanged)