A ray of light enters a glass slab of thickness 10 cm and refractive index 1.5 at an angle of 45 degrees. What is the lateral shift produced by the slab?

B. 10(1 - sqrt(2/3)) cm — The lateral shift is given by the formula d = t * sin(i - r) / cos(r). Using Snell's law, find the angle of refraction r and substitute the values.

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$A ray of light enters a glass slab of thickness and refractive index at an angle of incidence . What is the lateral shift produced by the slab?$

A. $10(1 - 1/sqrt(2)) cm$
B. $10(1 - sqrt(2/3)) cm$
C. $5(1 - sqrt(2/3)) cm$
D. $10(1 - sqrt(3/2)) cm$

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Correct answer: B — $10(1 - sqrt(2/3)) cm$

Solution

Using d = \frac{t s i n ( i - r )}{c o s ( r )} and sin (r) = \frac{sin ( 4 5 ^{\circ} )}{1.5} = \frac{1}{1.5 2} = \frac{2}{3}, we calculate the shift.

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A ray of light enters a glass slab of thickness t=10 cm and refractive index μ=1.5 at an angle of incidence i=45∘. What is the lateral shift d produced by the slab?

Solution

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$A ray of light enters a glass slab of thickness and refractive index at an angle of incidence . What is the lateral shift produced by the slab?$