In U-V protection treatment of eyeglass a thin layer of transparent medium is applied over glass- If refractive index of layer is and that of glass is 3-2 then minimum thickness of layer required in - m to block U-V rays of wavelength 400 nm is Write upto two digits after the decimal point-

For destructive interference of the light that passes through, 2 μ t=n λ 2 ×4/3t=n 400 × 10^ 9 t=n × 400 × 10^ 9× 3/2 × 4 t_min=150 × 10^ 9 m nbsp;(n = 1 ...

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Standard XII

Physics

Apparent Depth

In U-V protec...

Question

In U-V protection treatment of eyeglass a thin layer of transparent medium is applied over glass. If refractive index of layer is $\frac{4}{3}$ and that of glass is $\frac{3}{2}$ then minimum thickness of layer required in $μ m$ to block U-V rays of wavelength 400 nm is (Write upto two digits after the decimal point.)

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Solution

For destructive interference of the light that passes through,

$2 μ t = n λ$

$2 \times \frac{4}{3} t = n 400 \times 10^{- 9}$

$t = \frac{n \times 400 \times 10^{- 9} \times 3}{2 \times 4}$

$t_{m i n} = 150 \times 10^{- 9} m$ (n = 1)

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In U-V protection treatment of eyeglass a thin layer of transparent medium is applied over glass. If refractive index of layer is 43 and that of glass is 32 then minimum thickness of layer required in μm to block U-V rays of wavelength 400 nm is (Write upto two digits after the decimal point.)

For destructive interference of the light that passes through, 2μt=nλ 2×43t=n 400×10−9 t=n×400×10−9×32×4 tmin=150×10−9 m (n = 1)

For destructive interference of the light that passes through,

$2 μ t = n λ$

$2 \times \frac{4}{3} t = n 400 \times 10^{- 9}$

$t = \frac{n \times 400 \times 10^{- 9} \times 3}{2 \times 4}$

$t_{m i n} = 150 \times 10^{- 9} m$ (n = 1)