An optical fiber of core refractive index of 1.50 is to be clad to ensure that TIR will take place for a light ray with critical angle 85°. What minimum refractive index is allowed for cladding?

Q.1 (b) An optical fiber of core refractive index of 1.50 is to be clad to ensure that TIR will take place for a light ray with critical angle 85°. What minimum refractive index is allowed for cladding?

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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Given: \(n_1=1.50\), \(\theta_c=85^\circ\).

Formula: \(\sin\theta_c=\dfrac{n_2}{n_1}\Rightarrow n_2=n_1\sin\theta_c\)

\(n_2=1.50\times\sin85^\circ=1.50\times0.9962=1.494\)

Refractive index of cladding \(n_2\approx1.494\)

This is the limiting value. For TIR at 85°, \(\theta_c\) must not exceed 85°, so the cladding index must satisfy \(n_2\le1.494\), with 1.494 as the boundary value asked for.

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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An optical fiber of core refractive index of 1.50 is to be clad to ensure that TIR will take place for a light ray with critical angle 85°. What minimum refractive index is allowed for cladding?

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