Question

In a thin prism of glass \(\left(a_{(n) g}=1.5\right)\) which of the following relation between the angle of minimum deviation \(\delta_{\mathrm{m}}\) and the angle of refraction \(\mathrm{r}\) will be correct? (A) \(\delta_{\mathrm{m}}=(\mathrm{r} / 2)\) (B) \(\left(\delta_{\mathrm{m}} / 2\right)=\mathrm{r}\) (C) \(\delta_{\mathrm{m}}=1.5 \mathrm{r}\) (D) \(\delta_{\mathrm{m}}=\mathrm{r}\)

Short answer

The correct relationship between the angle of minimum deviation \(\delta_{m}\) and the angle of refraction r in a thin prism of glass with a refractive index of 1.5 is: \[\left(\frac{\delta_{m}}{2}\right) = r\]. So the correct answer is (B).

Step-by-step solution

Recall the formula for the angle of deviation in a prism

We will work with the formula: \[ \delta = (n - 1)A, \] where \(\delta\) is the angle of deviation, \(n\) is the refractive index of the prism, and \(A\) is the prism angle.

Analyze angle of minimum deviation and relate it to the angle of deviation

Since we want to find the relationship between the angle of minimum deviation (\(\delta_{m}\)) and the angle of refraction (r), let's first express the angle of deviation in terms of the angle of refraction. For a thin prism, the angle of minimum deviation \(\delta_{m}\) occurs when the angle of incidence i and angle of emergence e are approximately equal to the angle of refraction r. Therefore, we can now write the formula for the angle of deviation when \(\delta = \delta_{m}\) as: \[ \delta_{m} = (i + e) - A\] Since i and e are approximately equal to r in this case: \[ \delta_{m} = 2r - A\]

Replace the refractive index in the formula

We are given that the refractive index of the glass is 1.5 (\(n = 1.5\)). Thus, we can rewrite the formula for the angle of deviation as: \[ \delta = (1.5 - 1)A\]

Compare the formulas for angle of deviation and minimum angle of deviation

Now we have \[ \delta = 0.5A\] And for the minimum deviation, we have \[ \delta_{m} = 2r - A\]

Solve for the relationship between the angle of minimum deviation and the angle of refraction

We want to find a relationship between \(\delta_{m}\) and r. In order to do that, let's first solve for A in the angle of deviation equation as follows: \[\delta = 0.5A\] \[A = 2 \delta \] Now, substitute this expression for A into the minimum angle of deviation equation: \[\delta_{m} = 2r - 2\delta\] Divide the equation by 2 to get the relationship between \(\delta_{m}\) and r: \[\left(\frac{\delta_{m}}{2}\right) = r\]

Conclusion

The correct relationship between the angle of minimum deviation \(\delta_{m}\) and the angle of refraction r in a thin prism of glass with a refractive index of 1.5 is: \[\left(\frac{\delta_{m}}{2}\right) = r\] So the correct answer is (B).

Key concepts

Angle of Minimum Deviation

In the study of prisms, understanding the angle of minimum deviation is essential. This angle, often denoted as \( \delta_m \), is the smallest angle through which light is deviated as it passes through the prism. The angle of minimum deviation is significant because it happens when the light path within the prism is symmetrical. In simple terms, this means the light enters and exits the prism at equal angles.

For a thin prism, the minimum deviation occurs when the angles of incidence and emergence (i.e., how the light goes in and comes out) are equal. The formula for the angle of minimum deviation in a thin prism is given by:

• \( \delta_m = (i + e) - A \)

where \(A\) is the prism angle. When there's a scenario of minimum deviation in a thin prism, we can express it as:

• \( \delta_m = 2r - A \)

where \(r\) is the angle of refraction. Understanding \( \delta_m \) is useful in determining the optical properties of the material used in the prism.

Angle of Refraction

The angle of refraction is a crucial concept when discussing light behavior in prisms. It describes the angle that a refracted ray makes with the normal, which is an imaginary line perpendicular to the surface where the light enters or exits the prism. This angle is influenced by the material of the prism and the angle at which light hits the prism.

In the context of thin prisms, like in the given exercise, the angle of refraction, \(r\), becomes particularly interesting because it's directly linked to the angle of minimum deviation. When light undergoes minimum deviation within the prism, the incidence angle \(i\) and the emergence angle \(e\) are approximately equal to \(r\) (the angle of refraction).

• The relationship can be derived as \( \delta_m = 2r - A \) when considered for minimum deviation.

Grasping the concept of refraction allows us to better predict how different materials will influence the behavior of light as it passes through them, based on the refractive index which alters the value of \(r\).

Refractive Index

The refractive index, often symbolized as \( n \), is a measure that indicates how much a ray of light bends when it enters a material from another medium. It is a crucial factor in optics, helping to determine how light behaves when transitioning between different substances.

For instance, a refractive index of 1.5, like the glass prism in the exercise, means that light travels 1.5 times slower in the glass than in a vacuum. The refractive index is related to the angle of refraction, using Snell's Law, where \( n \) is given by the ratio of the sine of the angle of incidence to the sine of the angle of refraction:

• Snell's Law: \( n = \frac{\sin i}{\sin r} \)

In the case of thin prisms, where the aim is to determine a relationship between the angle of minimum deviation \( \delta_m \) and \( r \), a high refractive index indicates a larger bending effect. This relationship is simplified to \( \left(\delta_m / 2\right) = r \) for the exercise's specific condition. Understanding \( n \) is vital as it helps us ascertain the light's speed changes and how much it will bend.