Question

Can anyone help me with this question-22

Answer 1

Wavelength of light in air = 6000A°= 6000A°×10^-8 cm/A° = 6× 10^-5 cm

The velocity of light in air= 3×10^10 cm/s

Frequency v of light in air (velocity of light in air in cm/s) ÷ wave length of light in cm= ( 3×10^10 cm/s)/6×10^-5 cm= 5×10^14 Hz.

On entering the medium of refractive index n= 1.5 from air, the velocity of light decreases to c'= c/n= 3×10^10 cm per sec./ 1.5 = 2×10^10 cm/s.

The frequency of light does not change on refraction. So frequency of light in the medium= 5×10^14 Hz.

To keep the frequency unchanged, the wavelength of light in the medium also decreases by the same factor as the decrease in the velocity of light. So altered wavelength of light in the medium= 6000A°/1.5 = 4000A° .

So frequency of light in the medium= 5×10^14 Hz.

Wavelength of light in the medium= 4000A

HOPE IT HELPS YOU '_'

Answer 2

x_cm = (m₁x₁ + m₂x₂ + m₃x₃)/(m₁ + m₂ + m₃)
= [(1 × 2) + (2 × 0) + (3 × 2)] / (1 + 2 + 3)
= 8/6 m
= 4/3 m

y_cm = (m₁y₁ + m₂y₂ + m₃y₃)/(m₁ + m₂ + m₃)
= [(1 × 3) + (2 × 6) + (3 × 0)] / (1 + 2 + 3)
= 15/6 m

z_cm = (m₁z₁ + m₂z₂ + m₃z₃)/(m₁ + m₂ + m₃)
= [(1 × 0) + (2 × 3) + (3 × 1)] / (1 + 2 + 3)
= 9/6 m
= 3/2 m

∴ Centre of mass is at (4/3i + 15/6j + 3/2k)