Question

What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness \( h \) in terms of the angle of incidence \( i \) and angle of refraction \( r \), if the glass slab is placed in air medium?

• A. \( \frac{h \, \tan(i - r)}{\tan r} \)
• B. \( \frac{h \, \cos(i - r)}{\sin r} \)
• C. \( h \)
• D. \( \frac{h \, \sin(i - r)}{\cos r} \)

Hint:

The lateral shift is affected by the thickness of the glass slab and the angles at which the light enters and exits the slab. Understanding the geometry of refraction in a parallel-sided slab is essential for deriving the shift formula.

Answer 1

The Correct Option is D

To find the lateral shift \(d\) of a ray refracted through a parallel-sided glass slab of thickness \(h\), we use the geometry of refraction. When a light ray enters a glass slab with an angle of incidence \(i\) and refracts at an angle \(r\), the path of the light ray inside the slab creates a lateral shift.

The lateral shift \(d\) is given by:

\(d = \frac{h \sin(i-r)}{\cos r}\)

Here's how we derive the formula:

1. The refracted ray travels a horizontal distance \(L\) inside the slab, which can be given by \(L = h \sec r\), since \(h\) is the perpendicular distance (thickness of the slab) and \(r\) is the angle of refraction.
2. The lateral shift \(d\) is the horizontal component of this path difference, which can be expressed as \(L \sin(i - r)\).
3. Substitute \(L = h \sec r\) into the expression for the lateral shift:

\(d = h \sec r \cdot \sin(i-r) = \frac{h \sin(i-r)}{\cos r}\)

Therefore, the correct answer is \(\frac{h \, \sin(i - r)}{\cos r}\).