Find the minimum thickness of a layer of magnesium fluoride (n = 1.38) on flint glass (n = 1.66) that will cause destructive interference of reflected light of wavelength 5.50 xx 102 nm near the middle of the visible spectrum.

kjukks1234531

kjukks1234531

Answered question

2022-09-23

Nonreflective coatings on camera lenses reduce the loss of light at the surfaces of multilens systems and prevent internal reflections that might mar the image. Find the minimum thickness of a layer of magnesium fluoride (n = 1.38) on flint glass (n = 1.66) that will cause destructive interference of reflected light of wavelength 5.50 × 102   n m near the middle of the visible spectrum.

Answer & Explanation

Emaidedip6g

Emaidedip6g

Beginner2022-09-24Added 11 answers

Reflactive index of magnesium fluoride n m f = 1.38
Reflactive index of flint glass n f g = 1.66
Wavelength λ = 5.50 × 10 2   n m = 5.50 × 10 7   m
By the condition of destructive interference
2 n m f t = ( m + 1 2 ) λ t = ( m + 1 2 ) λ 2 n m f
For minimum thickness m=0
t m i n = 5.50 × 10 7 4 × 1.38 t m i n = 0.99637 × 10 7 t m i n = 99.637 × 10 9 t m i n = 99.64   n m
Answer: the minimum thickness of a layer of magnesium fluoride is equal to the 99.64 nm

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Atomic Physics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?