Answer:
The correct option is H
Explanation:
From the question we are told that
The index of refraction of coating is
The index of refraction of material is 
Generally the condition for constructive for a thin film interference is mathematically represented
![2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }](https://tex.z-dn.net/?f=2%20%2A%20%20t%20%20%3D%20%5B%20m%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5D%20%5Cfrac%7B%5Clambda%7D%7Bn_1%20%7D)
Here t represents the thickness
For minimum thickness m = 0
So

=> 
Decrease the amount of work done.
The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

- From this, the value of weight will be,

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
brainly.com/question/12830237
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Answer:
Explanation:
Image of distant object will be made at far point or at 52.5 so
object distance u = infinity
image distance v = - 52.5 cm
focal length required = f
Lens formula
1 / v - 1 / u = 1 / f
1 / - 52.5 - 0 = 1 / f
f = -52.5 cm
= -.525 m
Power P = 1 / f = - 1 / .525
= - 1.90
now , for eye with glass we shall find new near point .
v = ?
u = - 17.2 cm
f = - 52.5 cm
1 / v - 1 / u = 1 / f
1 / v + 1 / 17.2 = - 1 / 52.5
1 / v = - 1 / 17.2 - 1 / 52.5
= - .05813 - .019
= - .07713
u = - 12.96 cm
so new near point will be 12.96 cm