If the refractive index of glass is 1.5, what is the critical angle at the air-glass interface?
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Correct Answer: Option B
Explanation:
sin ic \(\frac{1}{n}\) = \(\frac{1}{1 - 5}\)
i = sin-1\(\frac{1}{1 - 5}\)
= sin-1\(\frac{10}{15}\)
= sin-1\(\frac{2}{3}\)
sin ic \(\frac{1}{n}\) = \(\frac{1}{1 - 5}\)
i = sin-1\(\frac{1}{1 - 5}\)
= sin-1\(\frac{10}{15}\)
= sin-1\(\frac{2}{3}\)