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Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an index of refraction of n= 1.50 and a glass plate with an index of n= 1.80, the diameter of the third bright ring is 0.700 mm .If water = (1.33) now fills the space between the lens and the glass plate, what is the new diameter of this ring? Assume the radius of curvature of the lens is much greater than the wavelength of the light

Answer :

ajeigbeibraheem

Answer:

the new diameter of the third ring = 0.607 mm

Explanation:

Consider the radius of [tex]\\m ^{th}\\[/tex] bright ring when air is in between the lens and the plate ;

[tex]r = \sqrt {\frac{(2m+1)\lambda R}{2}}[/tex]

Using the expression: [tex]r (n)= \sqrt {\frac{(2m+1)\lambda R}{2n}}[/tex] for the radius of the  [tex]\\m ^{th}\\[/tex] bright ring since the liquid (water ) of the refractive index 'n' is now in between the lens and the plate;

where;

[tex]\\m ^{th}\\[/tex] = number of fringe

λ = wavelength

R = radius

n = refractive index of water

Now ;

[tex]r (n)=\frac {r}{n}}[/tex]

the radius r of the third bright ring when the air is in between lens and plate = [tex]r= \frac{0.700 \ mm}{2} \\\\[/tex]

[tex]r= 0.35 \ mm[/tex]

The new radius of the third bright fringe is [tex]r(3) = \frac{0.35}{\sqrt 1.33}[/tex]

[tex]r(3) = 0.3035 \ mm[/tex]

Calculating the new diameter ; we have:

d(3) = 2(r(3))

d(3) = 2(0.3035)

d = 0.607 mm

Thus, the new diameter of the third ring = 0.607 mm

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