The incident angle ua shown in Fig. P33.53 is chosen so that the light passes symmetrically thro

step 1 For part A, we can write a snail law, that is N .A. Sine, sine, sine, theta, a, that is equal to

The incident angle ua shown in Fig. P33.53 is chosen so that...

The incident angle ua shown in $\textbf{Fig. P33.53}$ is chosen so that the light passes symmetrically through the prism, which has refractive index $n$ and apex angle $A$. (a) Show that the angle of deviation $\delta$ (the angle between the initial and final directions of the ray) is given by $$ \mathrm{sin} {A + \delta \over 2} = n \, \mathrm{sin} {A \over 2}$$ (When the light passes through symmetrically, as shown, the angle of deviation is a minimum.) (b) Use the result of part (a) to find the angle of deviation for a ray of light passing symmetrically through a prism having three equal angles ($A$ = 60.0$^\circ$) and $n$ = 1.52. (c) A certain glass has a refractive index of 1.61 for red light (700 nm) and 1.66 for violet light (400 nm). If both colors pass through symmetrically, as described in part (a), and if $A$ = 60.0$^\circ$, find the difference between the angles of deviation for the two colors.

Key Concepts

Geometry of Prisms

The geometry of a prism, particularly the relationship between its apex angle and the path of light through it, is central to understanding how light is deviated. The angles inside the prism are linked to the external angles by the prism's physical shape. This geometric insight is essential to deriving the relationship for the angle of deviation in terms of the prism's apex angle and its refractive index.

Dispersion

Dispersion is the phenomenon where a material’s refractive index varies with the wavelength of light. This variation causes light of different colors to be refracted by different amounts when passing through a prism, leading to a separation of colors. It is a key concept for explaining why, in applications such as spectroscopy, different wavelengths (such as red and violet light) are deviated by distinct angles.

Snell's Law

Snell's law is the fundamental principle that relates the angle of incidence and the angle of refraction when light passes from one medium to another. It states that n? sin(??) = n? sin(??), where n is the refractive index of the medium. This law is crucial in determining how much a light ray bends at an interface, which directly affects the analysis of light passing through a prism.

Minimum Deviation Condition

The minimum deviation condition occurs when the light passes symmetrically through a prism. Under this condition, the angles of incidence and emergence are equal, and the ray inside the prism is parallel to the base of the prism. This simplifies the geometric relationships, allowing for a derived expression that connects the prism's apex angle, the refractive index, and the angle of deviation. It also corresponds to the condition where the deviation of the ray is minimized.

Transcript

00:01 For part a, we can write a snail law, that is n .a.

00:06 Sine, sine, sine, theta, a, that is equal to nb, sine, satab.

00:27 Then we are given that a b is equal to a over 2, capital a over 2, capital a over 2.

00:37 So we can write this snail's law as sine, sine theta a, theta a is equal to nb, sine putting the value of theta bay, which is a over two, a over two here.

01:00 And then we know that a is equal to a plus 2, 2 alpha, alpha over 2.

01:16 So this is all over a plus 2 alpha.

01:20 Then we can plug the value of a theta a here and then our above equation becomes sign, sign a plus 2 alpha, a plus 2 alpha a plus 2 alpha over 2 is equal to so n and b will write n only and then this becomes a sign sign a over 2 a over 2 at each face of the prism the angle of deviation alpha is alpha so so for each phase then for two faces this is 2 alpha is equal to delta then we can replace 2 alpha with the delta in above equation then we can write equation the above equation will become sign a plus delta over 2 is equal to n sine sine a over 2 hence it is proved that a plus delta over 2 is equal to n times sine a over 2 in part b we are asked to find a delta for the given values of refractor index then we can solve for a delta from this equation and once we solved for the delta then the equation here we get here is a 2 times sine of inverse sine of inverse into refractive index sign of sign of a or 2 a or 2 a or 2 minus a so from a given data then we can plug those values and find the delta here so the delta will become a sine inverse inverse of refractory index is 1 .52 5 2 and sign of 60, sign of 60 over 2 that is 30, and here we have angle a is 60.

03:54 Then we can just do the maths and solve...

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