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Example · Example 8

Q.Calculate the critical angle for a glass-air interface, given the refractive index of the glass is 1.51.5.

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For a ray travelling from glass (n=1.5n=1.5) towards air (nair≈1n_{\text{air}}\approx1), the critical angle satisfies nsin⁡C=1×sin⁡90∘=1n\sin C=1\times\sin90^{\circ}=1, so sin⁡C=1n=11.5=0.6667\sin C=\dfrac{1}{n}=\dfrac{1}{1.5}=0.6667. Taking the inverse sine, C=sin⁡−1(0.6667)≈41.8∘C=\sin^{-1}(0.6667)\approx41.8^{\circ}. Any ray inside the glass striking the glass-air boundary at more than about …

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