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 / d_{0} + 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 10^{8} m/s) / (1.5 x 10^{8}
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)**

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