Three right angled prisms of refractive indices $n_{1}, n_{2}$ and $n_{3}$ are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then

n_{1} = n_{2} = n_{3}

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n_{1} = n_{2} \neq n_{3}

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1 + n_{1} = n_{2} + n_{3}

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1 + n_{2}^{2} = n_{1}^{2} + n_{3}^{2}

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Solution

The correct option is D $1 + n_{2}^{2} = n_{1}^{2} + n_{3}^{2}$
Applying Snell's Law at each surface,

AT B, $s i n i = n_{1} s i n r_{1} ⟹ s i n^{2} i = n_{1}^{2} s i n^{2} r_{1}$

At C, $n_{1} s i n (90 - r_{1}) = n_{2} s i n r_{2} ⟹ n_{1}^{2} c o s^{2} r_{1} = n_{2}^{2} s i n^{2} r_{2}$

At D, $n_{2} s i n (90 - r_{2}) = n_{3} s i n r_{3} ⟹ n_{2}^{2} c o s^{2} r_{2} = n_{3}^{2} s i n^{2} r_{3}$

At E, $n_{3} s i n (90 - r_{3}) = s i n (90 - i) ⟹ n_{3}^{2} c o s^{2} r_{3} = c o s^{2} i$

Adding all above equations gives,

$1 + n_{2}^{2} = n_{1}^{2} + n_{3}^{2}$

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Three right angled prisms of refractive indices n1,n2 and n3 are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then

Three right angled prisms of refractive indices $n_{1}, n_{2}$ and $n_{3}$ are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then