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For each polariser the angle given is the one for which light is transmitted and is given clockwise from the vertical. Horizontally polarised light is shone on a polarise that is angled at 35^circ to the vertical. If th e incoming light has amplitude 200Vm^-1 and intensity 53Wm^-2 work out: Part A Amplitude a) the amplitude of the transmitted light;

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For each polariser the angle given is the one for which light is transmitted and is given clockwise from the vertical. Horizontally polarised light is shone on a polarise that is angled at 35^circ to the vertical. If th e incoming light has amplitude 200Vm^-1 and intensity 53Wm^-2 work out: Part A Amplitude a) the amplitude of the transmitted light;

For each polariser the angle given is the one for which light is transmitted and is given clockwise from the vertical. Horizontally polarised light is shone on a polarise that is angled at 35^circ to the vertical. If th e incoming light has amplitude 200Vm^-1 and intensity 53Wm^-2 work out: Part A Amplitude a) the amplitude of the transmitted light;

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CliffordProfessional · Tutor for 6 years

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<br />Let's deal with the given question step by step while keeping clarity and conciseness at the forefront.<br /><br />The question is asking for the amplitude of the transmitted light, when horizontally polarised light is passed through a polarizer angled 35 degrees (clockwise from vertical) to the vertical axis. <br /><br />We've been given:<br />1. The angle theta ($\theta$) = 35 degrees <br />2. The initial amplitude (Ao) = 200 V/m<br />3. The initial intensity (Io) = 53 W/m^2<br /><br />Firstly, we need to understand Polarisation of Light. One of the ways light wave vibration is expressed is through polarisation. Transverse waves oscillate in multiple directions which includes up-down and side to side motion. However, in case of Polarisation, the light wave oscillates only in one direction.<br /><br />In this case, the light being horizontally polarised is coloured at 35º to the vertical plane. So, it's normal to the polarisation axis of the polarizer,(angled 35 degrees), titled towards the incoming light.<br /><br />The amplitude of polarised light is given by the following equation:<br /><br />A = Ao * cos(theta)<br /><br />Using the equation above,<br />Amplitude A = 200 * cos(35)<br />Calculating cos (35), we obtain a value approximately equal to 0.819152.<br />Therefore, Amplitude, A = 200 * 0.819152 = 163.83 V/m<br /><br />Hence, the amplitude of the transmitted light after crossing the 35-degree angled polarizer is about 163.83 V/m. This is because the movement of light becomes attenuated to a single plane after polarisation. I hope this explanation was satisfyingly clear, concise, and straight forward.<br /><br />【Answer】: The amplitude of the transmitted light is approximately 163.83 V/m.
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