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The drawing shows a rectangular block of glass ($n=1.52$) surrounded by liquid carbon disulphide ($n=1.63$). A ray of light is incident on the glass at point A with a $30{}^\circ $ angle of incidence. At what angle of refraction does the ray leave the glass at point B?

You are trying to photograph a bird sitting on a tree branch, but a tall hedge is blocking your view. However, as drawing shows, a plane mirror reflects light from the bird into your camera. For what distance must you set the focus of the camera lens in order to snap a sharp picture of the bird's image.

The drawing shows a horizontal beam of light that is incident on a prism made from ice. The base of the prism is also horizontal. The prism ($n=1.31$) is surrounded by oil whose index of refraction is $1.51$. Determine the angle $\theta $ that the exiting light makes with the normal to the right face of the prism.

A spotlight on a boat is $y=2.7\,\mathrm{m}$ above the water ($n=1.33$), and the light strike the water at a point that is $x=9.4\, \mathrm{m}$ horizontally displaced from the spotlight (see the drawing). The depth of the water is $4.0\, \mathrm{m}$. Determine the distance $d$, which locates the point where the light strikes the bottom from the boat.

A point source of light is submerged $2.2\,\mathrm{m}$ below the surface of a lake($n=1.33$) and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum area that this circle could have?

The drawing shows a rectangular block of glass ($n=1.52$) surrounded by liquid carbon disulfide ($n=1.30$). A ray of light is incident on the glass at point A with an angle of incidence of $30{}^\circ $ . At what angle of refraction does the ray leave the glass at point B?

A narrow beam of ultrasonic waves reflects off a liver tumor as shown. If the speed of the wave is $20\%$ less in the liver than in the surrounding tissues, determine the depth of the tumor.

A ray of light impinges from air onto a block of ice ($n=1.309$) at a $70{}^\circ $ angle of incidence. Assuming that this angle remains the same, find the difference ${\theta }_{2,ice}-{\theta }_{2,water}$ in the angles of refraction when the ice turns to water ($n=1.333$)

A horizontal beam of light is incident on face A of the prism ($n_2\ =\ 1.48$), which is surrounded by water ($n_1\ =\ 1.33$), as shown in the drawing. Find the angle that the beam emerges from the prism with respect to the normal.

A piece of ice is cut into a square block as shown. Its index of refraction is $1.3$. Light is incident on the top surface from the right as shown. What incident angle ${\theta }_c$ would lead to total internal reflection in the ice block, so no light would escape to the left?

A rectangular slab of glass has an index of refraction for red light ($\lambda =700\ \mathrm{nm}$) of $1.724$ and for blue light ($\lambda =450\ \mathrm{nm}$) of $1.770$. The angle of refraction for the blue beam inside the glass is $20.24{}^\circ $. What is the angle of refraction for the red beam?

A deep-sea diver wearing a spherical plastic helmet of diameter $0.50\, \mathrm{m}$ looks directly at a fish, and the fish looks directly back. The fish is $2.00\, \mathrm{m}$ from the helmet, and the diver's eye is $0.20\, \mathrm{m}$ from the helmet. Ignore any refraction effects from the plastic helmet itself. The index of refraction of the water is $1.33$.

(a) How far does the fish appear to be from the surface of the helmet as observed by the diver?

(b) How far does the diver's eye appear to be from the surface of the helmet as observed by the fish?

(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

\[\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{+R}\]

\[\frac{1.33}{2}+\frac{1}{q}=\frac{1-1.33}{0.5/2}\]

\[\Rightarrow \frac{1}{q}=-\frac{0.33}{\frac{0.5}{2}}-\frac{1.33}{2}=1.985\Rightarrow q=-0.503\ \mathrm{m}\]

The sign convention for refraction cases is 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 $R<0$.

\[\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{-R}\]

\[\frac{1}{0.2}+\frac{1.33}{q}=\frac{1.33-1}{-\left(\frac{0.5}{2}\right)}\Rightarrow q=-0.21\ \mathrm{m}\]

The negative sign of $q$ means that the image of the diver's eye is on the same side of the object.

The relation $n_1{\,\sin {\theta }_1\ }=n_2{\,\sin {\theta }_2\ }$ which applies as a ray of light strikes an interface between two media, is known as:

(a) Gauss's law

(b) Snell's law

(c) Faraday's law

(d) Cole's law

(e) Law of sines

The index of refraction of benzene is $1.80$. The critical angle for total internal reflection, at a benzene-air interface, is about:

(a) $56{}^\circ $

(b) $47{}^\circ $

(c) $34{}^\circ $

(d) $22{}^\circ $

(e) $18{}^\circ $

When light passes from a medium of high index of refraction to one of lower index of refraction a physical phenomenon called total internal reflection occurs. The critical angle ${\theta }_c$ for which this phenomenon occurs at the interface of the mediums is found as

\[{\,\sin {\theta }_c\ }=\frac{n_2}{n_1}\ ,\ \ (n_1>n_2)\]

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

\[{\,\sin {\theta }_c\ }=\frac{n_2}{n_1}=\frac{1}{1.8}\Rightarrow \ {\theta }_c=33.7{}^\circ \]

The correct answer is C.

Monochromatic light, at normal incidence, strikes a thin film in air. If $\lambda $ denotes the wavelength in the film, what is the thinnest film in which the reflected light will be a maximum?

(a) Much less than $\lambda $

(b) $\lambda /4$

(c) $\lambda /2$

(d) $3\lambda /4$

(e) $\lambda $

Category : Refraction

MOST USEFUL FORMULA IN REFRACTION:

Law of refraction (Snell's law):

\[n_1\sin \theta_i=n_2\sin \theta_r\]

where $n$s are the indexes of refraction of the materials.

$\theta_i$ and $\theta_r$ are the angles of incidence and refraction.

Total internal reflection:

when light travells from a medium with high refractive index to that with low refractive index this phenomena will occur. such as water to air.

Critical angle is the angle at which total internal reflection can be occured

\[\sin \theta_{cri}=\frac{n_b}{n_a} \quad , \quad n_a>n_b\]

If the angle of incidence $\theta_a$ in the dense medium is larger than or equal to $\theta_{cri}$ then total internal reflection will occur.

Number Of Questions : 15

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