Solved Problems

Snell's law: describes how light bends when traveling from one medium to the next. i.e. it states that $n_1\,{\sin {\theta }_1\ }=n_2\,{\sin {\theta }_2\ }$ , where $\theta $'s represents the angles the light makes with the normal in the mediums and $n$ is the refractive index of the object or medium. 

Now first use the Snell's law and find the refracted angle ${\theta }_1$ from the left side
\begin{align*} n\,{\sin {\theta }_i\ }&=n^{'}\,{\sin {\theta }_1\ } \\ 1.63\,{\sin 30{}^\circ \ }&=1.52\,{\sin {\theta }_1\ } \\ \to {\theta }_1&=32.42{}^\circ \end{align*} 
From the geometry and the drawing above, we have \[{\theta }_1+{\theta }_2=90\Rightarrow \ {\theta }_2=90-32.42=57.58{}^\circ\]
By applying the Snell's law again at point B, we obtain 
\begin{align*} n^{'}\,{\sin {\theta }_2\ }&=n\,{\sin {\theta }_r\ } \\ \to {\theta }_r&={{\sin}^{-1} \left(\frac{n^{'}}{n}{\sin {\theta }_2\ }\right)\ }\\ &=51.92\end{align*}

Solution 1: In plane mirrors, the angle of incidence ${\theta }_i$ equals the angle of reflection ${\theta }_r$ i.e. ${\theta }_i={\theta }_r$.

In order to get a sharp picture we must determine $h_1+h_2$ i.e. the direct distance of image from camera. As drawing shows 
\[{\tan {\theta }_r\ }=\frac{x}{3.7}={\tan {\theta }_i\ }=\frac{4.3-x}{2.1}\ \ \ \Rightarrow \ \ x=\frac{3.7}{2.1}\left(4.3-x\right)=2.74\mathrm{m}\] 

using the Pythagorean theorem, we can obtain the values of $h_1$ and $h_2$

\begin{gather*} h_1=\sqrt{x^2+{\left(3.7\right)}^2}=4.60 \\ h_2=\sqrt{{\left(4.3-x\right)}^2+{\left(2.1\right)}^2}=2.61 \\ \therefore\  h_1+h_2=7.21\ \mathrm{m}\end{gather*}
Solution 2: plane mirrors forms a virtual image! Therefore, camera sees a virtual image and the blue triangle in the drawing shows this image (using the Pythagorean Theorem)
\[d=\sqrt{{\left(3.7+2.1\right)}^2+{\left(4.3\right)}^2}=7.21\ \mathrm{m}\] 
 

Use the ray diagram in the below and apply the Snell's law to determine $\theta $:
\[n_1\,{\sin 27{}^\circ \ }=n_2\,{\sin \alpha \ }\to \alpha ={{\sin}^{-1} \left(\frac{n_1}{n_2}{\sin 27{}^\circ \ }\right)\ }=31.55{}^\circ \] 
From the geometry we have (the sum of the internal angles of a triangle is $180{}^\circ $): 
\[\left(90-\alpha \right)+54+\left(90-\beta \right)=180{}^\circ \to \ \alpha +\beta =54\ \Longrightarrow \beta =22.45{}^\circ \] 
\[n_2\,{\sin \beta \ }=n_1\,{\sin \theta \ }\to \theta ={{\sin}^{-1} \left(\frac{n_2}{n_1}{\sin \beta \ }\right)\ }=19.34{}^\circ \] 

This is just a simple combination of geometry and Snell's law:
\[n\,{\sin {\theta }_i\ }=n_w\,{\sin {\theta }_r\ }\] \[\Rightarrow {\theta }_r={{\sin}^{-1} \left(\frac{n}{n_w}{\sin {\theta }_i\ }\right)\ }={{\sin}^{-1} \left(\frac{1}{1.33}\frac{9.4}{\sqrt{{9.4}^2+{2.7}^2}}\right)\ }=46.27{}^\circ \] using the definition of the Tangent of an angle in a right angled triangle, we can find the desired distance $d$ as follows: 
\[{\tan {\theta }_r\ }=\frac{d-x}{4}\ \ \Rightarrow \ \ d=4\times {\tan 46.27{}^\circ \ }+x=4.18+9.4=13.58\ \mathrm{m}\] 

All the rays emitted from the point source, are refracted at various angles. If the incident angle ${\theta }_i$ is such that the refracted angle becomes $90{}^\circ $, then all of the energy of the ray transmitted parallel to the surface. This incident angle ${\theta }_c$ is called the critical angle.

All the rays with an incident angle greater than ${\theta }_c$, will be reflected back into the medium from which they came (total reflection).

Therefore, we must find the critical angle and then from its geometry find the maximum area that forms.
\[{\sin {\theta }_c\ }=\frac{n_1}{n_2}=\frac{1}{1.33}\to {\theta }_c=48.75{}^\circ \] 
\[{\tan {\theta }_c\ }=\frac{r}{2.2}\to r=2.2\,{\tan 48.75{}^\circ \ }=2.50\,\mathrm{m}\] 
\[Area=\pi r^2=\pi {\left(2.50\right)}^2=19.77\ {\mathrm{m}}^{\mathrm{2}}\] 
 

Let $n_1=1.30$ and $n_2=1.51$

Use Snell's law to find the refracted light at  point A \[n_1\,{\sin 30{}^\circ \ }=n_2\,{\sin {\theta }_2\ }\to {\theta }_2=25.5{}^\circ \]
From the geometry, we have: \[{\theta }_2+{\theta }_3=90\to {\theta }_3=90-25.5=64.5{}^\circ \]
Now because the light passes from a higher refractive index ($n_2$) into a medium with a lower refractive index ($n_1$) we must first determine the critical angle.

If the incident angle is greater than ${\theta }_c$, then total internal reflection occurs. 
The critical angle is found by the following formula: \[{\sin {\theta }_c\ }=\frac{n_1}{n_2}=\frac{1.30}{1.51}\to {\theta }_c=59.42{}^\circ \]
Because ${\theta }_3>{\theta }_c$ so total internal reflection occurs.

Light leaves B at angle ${\theta }_3$ and as shown in the figure, after occurring total internal reflection once again at the bottom side, the light ray leaves the block at angle ${\theta }_R$.

Therefore, use again the Snell's law at right side
\begin{align*} n_2\,{\sin {\theta }_2\ }&=n_1\,{\sin {\theta }_R\ }\\ \\ \to {\theta }_R&={{\sin}^{-1} \left(\frac{n_2}{n_1}{\sin {\theta }_2\ }\right)\ }\\ \\&={{\sin}^{-1} \left(\frac{1.51}{1.30}{\sin 25.5{}^\circ \ }\right)\ }\\ \\&=30{}^\circ \end{align*} 

Note: the refractive index defines as 
\[n=\frac{c}{v}\] 
Where $c$ is the speed of light in the vacuum (which is constant) and $v$ is the velocity of light (or other waves) in the medium. So 
From Snell's law:
\begin{align*} n_{tiss}\,{\sin 50\ }&=n_{liv}\,{\sin \alpha \ } \\ \\ \to \alpha &={{\sin}^{-1} \left(\frac{n_{tiss}}{n_{liv}}{\sin 50{}^\circ \ }\right)\ }\end{align*} 
Since the refractive index of the mediums are not given so use the definition of the refractive index to find the ratio of the refractive index of the mediums in terms of velocities as follows
\[\frac{n_{tiss}}{n_{liv}}=\frac{v_{liv}}{v_{tiss}}=0.8\] 
Now substitute it into $\alpha $
\[\alpha ={{\sin}^{-1} \left(0.8{\sin 50{}^\circ \ }\right)\ }=37.79{}^\circ \] 
From the law of reflection, $w=\frac{12}{2}=6\ \mathrm{cm}$
\[\therefore {\tan \alpha \ }=\frac{w}{d}\ \Longrightarrow d=\frac{6}{{\tan 37.79{}^\circ \ }}=7.73\ \mathrm{cm}\] 
 

Use Snell's law to find the refracted ray in the second medium: \[n_1\,{\sin {\theta }_1\ }=n_2\,{\sin {\theta }_2\ }\to {\theta }_2={{\sin}^{-1} \left(\frac{n_1}{n_2}{\sin {\theta }_1\ }\right)\ }\mathrm{but}\ n_1=1\ \]  Now substitute the values of the ice and water into the above relation and then subtract them from each other \begin{align*} \Rightarrow {\theta }_{2,ice}-{\theta }_{2,water} &={{\sin}^{-1} \left(\frac{{\sin {\theta }_1\ }}{n_{ice}}\right)-\ }{{\sin}^{-1} \left(\frac{{\sin {\theta }_1\ }}{n_{water}}\right)\ }\\ & ={{\sin}^{-1} \left(70{}^\circ \frac{1}{1.309}\right)\ }-{{\sin}^{-1} \left(70{}^\circ \frac{1}{1.333}\right)\ }=1.054{}^\circ \end{align*} 
 

Light is incident at angle $70{}^\circ$ on the face $C$. Because light moves through a denser medium into a less dens one, then we must first determine the critical angle $\theta_{cri}$ as below \[{\sin {\theta }_{cri}\ }=\frac{n_1}{n_2}\to {\theta }_{cri}={{\sin}^{-1} \left(\frac{1.33}{1.48}\right)\ }\sim \ 64{}^\circ \] Since ${\theta }_{rc}>{\theta }_{cri}$ therefore total internal reflection can occur.

From the geometry (colored triangle), we can find the incident angle ${\theta }_i=50{}^\circ $ on the face $B$ and then using the Snell's law we obtain the refracted ray on face $B$
\[n_2\,{\sin {\theta }_i\ }=n_1\,{\sin {\theta }_f\ }\Rightarrow {\theta }_f={{\sin}^{-1} \left(\frac{n_2}{n_1}{\sin 50^\circ \ }\right)\ }=58.5^\circ \] 
Note: the sum of the internal angles of a triangle is always $180^\circ $.

Recall that two conditions are required to occur total internal reflection:
(a) The light propagating from a more dense medium into a less dense medium.

(b) The angle of incidence is greater than the critical angle.

Use Snell's law formula to find the ${\theta }_c$
\[n_a\,{\sin {\theta }_c\ }=n_i\,{\sin {\theta }_1\ }\Rightarrow (1.3)\,{\sin {\theta }_1\ }=1\,{\sin {\theta }_c\ }\] 
Now calculate the critical angle for the ice-air boundary as \begin{align*}n_i\,{\sin {\theta }_{cri}\ }&=n_a\,{\sin 90{}^\circ \ }\\ \\ \Rightarrow {\sin {\theta }_{cri}\ }&=\frac{1}{1.3} \\ \\\Rightarrow {\theta }_{cri}\ \sim \ 50.28{}^\circ \end{align*}  (For critical angle, the angle of refraction is $90{}^\circ $).

Therefore, if ${\theta }_2>{\theta }_{cri}$ then the total internal reflection occurs on the left side.

By supposing ${\theta }_2={\theta }_{cri}$ and using the figure, we have
\[{\theta }_1+{\theta }_2=90{}^\circ \Rightarrow {\theta }_1=90-50.28\ \sim \ 39.72{}^\circ \] 
By substitute it into the following relation, we obtain the minimum angle that the total internal reflection can occur.
\begin{align*} (1.3)\,{\sin {\theta }_1\ }&=1\,{\sin {\theta }_c\ } \\ \Rightarrow {\theta }_c&={{\sin}^{-1} ((1.3){\sin 39.72^\circ})\ }\sim \ 56.17^\circ \end{align*} 
Therefore, if ${\theta }_c>56.17{}^\circ $, total internal reflection occurs.
 

First, use the Snell's law to find the incident angle as
\begin{align*} n_i\,{\sin {\theta }_i}&=n_b\,{\sin {\theta }_{rb}}\\ \\ \Rightarrow {\theta }_i&={{\sin}^{-1} \left(\frac{n_b}{n_i}{\sin {\theta }_{rb}\ }\right)\ }\\ \\ &={{\sin}^{-1} \left(\frac{1.770}{1}{\sin 20.24{}^\circ \ }\right)\ }\\ \\&=37.75{}^\circ \end{align*} Now using Snell's law once again, calculate the reflected angle for red beam as \begin{align*} n_i\,{\sin {\theta }_i\ }&=n_r\,{\sin {\theta }_{rr}\ } \\ \\ \Rightarrow {\theta }_{rr}&={{\sin}^{-1} \left(\frac{n_i}{n_r}{\sin {\theta }_i\ }\right)\ }\\ \\ &={{\sin}^{-1} \left(\frac{1}{1.724}{\sin 37.75{}^\circ \ }\right)\ }\\ \\&=20.80{}^\circ \end{align*}
 

(a) In optics, images can form by reflection from a spherical mirrors or refraction by a refracting surface (thin lenses,$\dots$). For reflection from a spherical mirror with focal length $f$, the image $q$ and object $p$ distances are related together by the following relation 
\[\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\] 
But the image and object distances from a spherical boundary of radius $R$ which is between the two transparent materials of indices of refraction $n_1$ and $n_2$ is given by the well-known equation below
\[\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}\] 
The sketch of the problem is shown.

Using the above equation, we obtain
\begin{gather*} \frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{+R} \\ \\  \frac{1.33}{2}+\frac{1}{q}=\frac{1-1.33}{0.5/2} \\ \\ \Rightarrow \frac{1}{q}=-\frac{0.33}{\frac{0.5}{2}}-\frac{1.33}{2}=1.985 \\ \\ \Rightarrow q=-0.503\quad {\rm m}\end{gather*}
The sign convention for refraction cases is the same as thin lenses. Since the center of curvature is on the refracted-light side of the surface so $R>0$.

The negative sign for $q$ indicates that the image is in front (incident-light side) of the helmet, in other words, it is in the same medium as the object.

Therefore, the image of the fish must be virtual and closer to the helmet than it actually is. 

(b) In this case since the center of curvature is located in incident-light side of the refracting surface so $R<0$. 
\begin{gather*} \frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{-R} \\ \\  \frac{1}{0.2}+\frac{1.33}{q}=\frac{1.33-1}{-\left(\frac{0.5}{2}\right)} \\ \\ \Rightarrow q=-0.21\quad {\rm m}\end{gather*} The negative sign of $q$ means that the image of the diver's eye is on the same side of the object.

Snell's law describes the bending of a light ray as it passes from one medium to another. In the mathematical language is $n_1{\,\sin {\theta }_1\ }=n_2{\,\sin {\theta }_2\ }$ where n's are the refractive indices. 

The correct answer is B. 
 

When light passes from a medium of high index of refraction to a medium of lower index of refraction, a physical phenomenon called total internal reflection occurs. The critical angle $\theta_c$ at which this phenomenon occurs at the interface of the mediums is found as \[\sin \theta_c=\frac{n_2}{n_1}\quad ,\quad (n_1>n_2)\] 
Benzene and air have indices of refraction $n_1=1.8\ ,\ n_2=1$, respectively. Thus the critical angle at which total internal reflection occurs when light travels from benzene to air is \begin{gather*} \sin \theta_c=\frac{n_2}{n_1}=\frac{1}{1.8} \\\\ \Rightarrow \quad \theta_c=33.7^\circ \end{gather*}

For more solved problems on the index of refractions refer, here.

When a monochromatic light incident on a thin film of refractive index $n$ and thickness $t$, due to combination of the reflected rays from upper and lower surfaces of the film interference in the reflected rays can be occurred. In general, the condition for constructive interference in thin films is 
\[2t=\left(m+\frac{1}{2}\right){\lambda }_n\ ,\ \ m=(0,1,2,\dots )\] 
Where ${\lambda }_n$ is the wavelength of light in the film which is found by ${\lambda }_n=\lambda /n$. $\lambda $ is the wavelength in the vacuum.
Maximum pattern in the interference language means a constructive fringe. $m=0$ in the equation above, gives the thinnest film that is
\[t=\frac{\left(m+\frac{1}{2}\right){\lambda }_n}{2}=\frac{{\lambda }_n}{4}\] 

The correct answer is B.