A glass cube having edge 4 cm is kept on a printed page. If the refractive index of glass material is 1.5, then by what amount will the printed letters appear to be shifted, while viewed from the top?

Question

A glass cube having edge 4 cm is kept on a printed page. If the refractive index of glass material is 1.5, then by what amount will the printed letters appear to be shifted, while viewed from the top? (A) 1.33 cm (B) 2.33 cm (C) 0.6 cm (D) 1.66 cm

Tardigrade Admin · Accepted Answer

The edge of the glass cube is 4 cm. Thus, the thickness, t = 4 cm The shift in position of printed letters is given by Δ t=( 1-(1/μ ) )t=( 1-(1/1.5) )× 4 =(4/3)=1.33cm

Q. A glass cube having edge 4 cm is kept on a printed page. If the refractive index of glass material is 1.5, then by what amount will the printed letters appear to be shifted, while viewed from the top?

1.33 cm

2.33 cm

0.6 cm

1.66 cm

The edge of the glass cube is 4 cm. Thus, the thickness, t = 4 cm The shift in position of printed letters is given by $Δ t = (1 - \frac{1}{μ}) t = (1 - \frac{1}{1.5}) \times 4$ $= \frac{4}{3} = 1.33 c m$

Q. A glass cube having edge 4 cm is kept on a printed page. If the refractive index of glass material is 1.5, then by what amount will the printed letters appear to be shifted, while viewed from the top?

1.33 cm

2.33 cm

0.6 cm

1.66 cm

The edge of the glass cube is 4 cm. Thus, the thickness, t = 4 cm The shift in position of printed letters is given by Δt=(1−μ1​)t=(1−1.51​)×4 =34​=1.33cm

The edge of the glass cube is 4 cm. Thus, the thickness, t = 4 cm The shift in position of printed letters is given by $Δ t = (1 - \frac{1}{μ}) t = (1 - \frac{1}{1.5}) \times 4$ $= \frac{4}{3} = 1.33 c m$