In the given diagram a wavefront AB moving in air is incident on a plane glass surface xy. Its position CD after refraction through the glass slab is shown also along with normal drawn at $A$ and $D$ \. The refractive index of glass will be equal to

(B D / A C)

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(A B / C D)

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(sin ϕ / sin ϕ^{'})

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(cos θ / cos θ^{'})

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Solution

The correct option is C $(sin ϕ / sin ϕ^{'})$
$\begin{matrix} t = \frac{V_{a}}{V_{g}} = \frac{B D}{A C} Δ A B D B D = A D sin θ Δ A C D A C = A D sin θ^{'} μ = \frac{sin θ}{sin θ^{'}} \end{matrix}$

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In the given diagram a wavefront AB moving in air is incident on a plane glass surface xy. Its position CD after refraction through the glass slab is shown also along with normal drawn at A and D\. The refractive index of glass will be equal to

The correct option is C (sinϕ/sinϕ′)t=VaVg=BDACΔABDBD=ADsinθΔACDAC=ADsinθ′μ=sinθsinθ′

In the given diagram a wavefront AB moving in air is incident on a plane glass surface xy. Its position CD after refraction through the glass slab is shown also along with normal drawn at $A$ and $D$ \. The refractive index of glass will be equal to

The correct option is C $(sin ϕ / sin ϕ^{'})$
$\begin{matrix} t = \frac{V_{a}}{V_{g}} = \frac{B D}{A C} Δ A B D B D = A D sin θ Δ A C D A C = A D sin θ^{'} μ = \frac{sin θ}{sin θ^{'}} \end{matrix}$