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Tutor profile: Neel C.

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Neel C.
Ph.D. Candidate at University of Toronto
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Questions

Subject: Physics (Waves and Optics)

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Question:

Find the critical angle condition for a light ray propagating from glass (refractive index = 1.5) to air (refractive index = 1) so that it is observes total internal reflection in the glass medium.

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Neel C.
Answer:

According to Snell's law $$n_i sin(\theta_i) = n_t sin(\theta_t)$$, where $$n_1$$ is the incident medium (here glass) and $$n_2$$ is the transmitted medium (here air). $$\theta_i$$ is the incident angle and $$\theta_t$$ is the transmitted angle, both measured with respect to the normal to the interface. At critical angle the transmitted angle $$\theta_t$$ = 90 degrees. Substituting this and the refractive indices in the Snell's law, we get the critical angle as: $$1.5\times \theta_i = 1\times sin(90)$$ Therefore, $$\theta_i = sin^{-1}(1/1.5)$$

Subject: Physics (Electricity and Magnetism)

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Question:

What is the impedance of a series LCR circuit in resonance? What is the resonant frequency and what is the phase angle at resonance?

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Neel C.
Answer:

Resonance is a condition in the LCR circuit when the output of the circuit is maximum for a particular frequency. In an LCR circuit, resonance occurs when the value of the inductive reactance ($$X_L$$) and capacitive reactance ($$X_C$$) is equal, but they have a phase difference of $$\pi$$ radians. Thus, they cancel each other. $$ Z = \sqrt{R^2+(X_L-X_C)^2}$$ where R is the resistance. As $$X_L = X_C$$, Z = R. Moreover, $$X_L = \omega L$$, and $$X_C = \frac{1}{\omega C}$$, so at resonance, $$\omega L = \frac{1}{\omega C}$$ $$\therefore \omega^2 = \frac{1}{LC}$$ $$\therefore \omega = \frac{1}{\sqrt{LC}}$$, but since $$f = \frac{\omega}{2\pi}$$, then $$f = \frac{1}{2\pi\sqrt{LC}}$$ This f is the resonant frequency of the series LCR circuit. Also, since the impedance of circuit is R, the phase of impedance is 0, as R is a real quantity.

Subject: Linear Algebra

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Question:

Show that the vectors $$ \textbf i = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$ , $$\textbf j = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} $$ and $$\textbf k = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} $$ form an orthonormal basis.

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Neel C.
Answer:

First, show that the inner product of each pair of vectors is equal to 0. $$\textbf i \cdot \textbf j = (1)(0)+(0)(1)+(0)(0) = 0$$ $$\textbf i \cdot \textbf k = (1)(0)+(0)(0)+(0)(1) = 0$$ $$\textbf j \cdot \textbf k = (0)(0)+(1)(0)+(0)(1) = 0 $$ Therefore the three vectors are orthogonal and also linearly independent. Then, show that the length or norm of each vector is equal to 1. $$||i|| = \sqrt {1^2+0^2+0^2} = \sqrt 1 = 1$$ $$||j|| = \sqrt {0^2+ 1^2+0^2} = \sqrt 1 = 1$$ $$||j|| = \sqrt {0^2+ 0^2+1^2} = \sqrt 1 = 1$$ The three vectors are orthogonal to each other and the length of each is equal to 1; therefore they form an orthonormal basis.

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