Physics: What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 566 nm light? The index of refraction of the coating material is 1.46 and the index of the glass is 1.71.

What is the minimum thickness of coating which should

starreee

30.01.2022

The thickness is $t = 1.415 *10^{-7 } \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 566 \ nm = 566 *10^{-9} \ m$

The index of refraction of glass is n_g = 1.71

The index of refraction of the coating is n= 1.46

Generally the condition for destructive interference is

$2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }$

Here m is the order of the interference pattern and given from the question that we are considering minimizing reflection m = 0

t = thickness of the coating

substituting values

$2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }$

=> $t = 1.415 *10^{-7 } \ m$

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Kurlyash

10.02.2023

I = 30 A.

Explanation:

Given that,

The voltage of the battery, V = 230 V

Power used to charge the battery, P = 6.9 kW

We need to find the current used to charge the battery. The formula for the power is given by :

P = VI

Where

I is current

So,

$I=\dfrac{P}{V}\\\\I=\dfrac{6.9\times 10^3}{230}\\\\=30\ A$

So, the required current is 30 A.

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