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| | | ### Statement |
| | | |
| | − | $13.2.14.$ [Insert problem description here] |
| | + | $13.2.14.$ a. A light beam passes through a wedge with a small angle $\alpha$ at the vertex, |
| | + | which is perpendicular to the front face of the wedge. Prove that the angle |
| | + | of deflection of the ray from the initial direction is approximately $(n − 1) \alpha$. |
| | + | Refractive index of the wedge $n$. |
| | + | b. A light ray passes through a wedge with a small angle $\alpha$ at the vertex, |
| | + | which falls at a small angle $\gamma$ to the normal to the front surface of the wedge. |
| | + | Prove that the angle of deflection of the light beam from the initial direction |
| | + | is approximately $(n − 1) \alpha$. Refractive index of the wedge $n$. |
| | + | c. Print the formula for the focal length of a thin lens. The radii of curvature |
| | + | of the lens surface are $R_1$ and $R_2$, and the refractive index of the lens material |
| | + | is $n$. |
| | | |
| | − | __Example Statement__: |
| | − | $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. |
| | + |  |
| | | |
| | − | |
| | | ### Solution |
| | | |
| | − | [Your solution should be placed here] |
| | + | #### 1. Hitting the first surface |
| | | |
| | − | __Example Solution__: |
| | − | The acceleration of the body defined by |
| | + | Let's imagine there is a parallel lays of light entering the lens with refractive index $n_1$ from media with refractive index $n_0$, and after lens there is a media with refractive index $n_2$. |
| | | |
| | − | $$a(t) = bt$$ |
| | + | First, let's consider the moment when ray hits front of lens at the point $h$ above optical axis. |
| | | |
| | − | We know that acceleration is the time derivative of velocity: |
| | + |  |
| | | |
| | − | $$a(t) = \frac{d v(t)}{d t}$$ |
| | + | As shown in the picture, the incidence angle is $$\tan(\alpha) \approx \frac{h}{R_1} \approx sin{(\alpha)} \approx \alpha$$ |
| | + | where $R_1$ is the curvature radius of first surface. |
| | | |
| | − | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: |
| | + | If there wasn't second surface, the ray after refraction will cross the optical axis at point $x_1$, and angle is $\tan(b) \approx \sin(\beta) \approx \beta = x_1/h$. So, the refraction angle is |
| | + | $$ \alpha - \beta = \frac{h}{R_1} - \frac{h}{x_1}$$ |
| | + | And by Snell's law: |
| | + | $$n_0 \cdot \frac{h}{R_1} = n_1 \cdot \left(\frac{h}{R_1} - \frac{h}{R_2}\right)$$ |
| | | |
| | − | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ |
| | + | #### 2. Hitting the second surface |
| | | |
| | − | If the initial velocity is $v(0) = 0$, then the velocity becomes: |
| | + | After hitting the second surface, the ray will cross the axis at point $x_2$, which we want to find. |
| | | |
| | − | $$v(t) = \frac{b t^2}{2}$$ |
| | + |  |
| | | |
| | − | Likewise, integrate $v(t)$ with respect to time: |
| | + | The incidence angle is |
| | + | $$ |
| | + | \alpha_2 = \frac{h}{R_2} + \frac{h}{x_1} |
| | + | $$ |
| | + | And the refraction angle is |
| | + | $$ |
| | + | \beta_2 = \frac{h}{R_2} + \frac{h}{x_2} |
| | + | $$ |
| | + | By Snell's Law: |
| | + | $$ |
| | + | n_1 \cdot \alpha_2 = n_2 \cdot \beta_2 |
| | + | $$ |
| | + | Therefore: |
| | + | $$ |
| | + | n_1 \cdot \left(\frac{h}{R_2} + \frac{h}{x_1} \right) = n_2 \cdot \left(\frac{h}{R_2} + \frac{h}{x_2} \right) |
| | + | $$ |
| | + | So, we have a system of equations: |
| | + | $$\begin{cases} |
| | + | n_0 \cdot \frac{h}{R_1} = n_1 \cdot \left(\frac{h}{R_1} - \frac{h}{R_2}\right)\\ |
| | + | n_1 \cdot \left(\frac{h}{R_2} + \frac{h}{x_1} \right) = n_2 \cdot \left(\frac{h}{R_2} + \frac{h}{x_2} \right) |
| | + | \end{cases} |
| | + | $$ |
| | + | Solving it, we can find $\frac{1}{x_2}$: |
| | + | $$ |
| | + | \frac{1}{x_2} = \left(\frac{n_1}{n_2} -1 \right) \cdot \frac{1}{R_2} + \left(\frac{n_1}{n_2} - \frac{n_0}{n_2} \right) \cdot \frac{1}{R_1} = \frac{1}{F} |
| | + | $$ |
| | + | In this problem, we have $n_0 = n_2 = 1, n_1 = n$ |
| | | |
| | − | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ |
| | + | So, we have |
| | + | $$ |
| | + | \boxed{\frac{1}{F} = \left(n-1 \right) \cdot \left(\frac{1}{R_1} + \frac{1}{R_2} \right)} |
| | + | $$ |
| | | |
| | − | From where the coordinate from time, considering the initial conditions: |
| | | |
| | − | $$\boxed{x(t)=\frac{bt^3}{6}}$$ |
| | − | |
| | | #### Answer |
| | | |
| | − | [Insert a concise answer or boxed result, like this:] |
| | − | |
| | − | |
| | − | __Example Answer__: |
| | − | $$ x(t)=\frac{bt^3}{6} $$ |
| | + | $$\frac{1}{F}=(n-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$$ |
