JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
A prism of refractive index \(n_{1}\) and another prism of refractive index \(n_{2}\) are stuck together (as shown in the figure). \(n_{1}\) and \(n_{2}\) depend on \(\lambda\), the wavelength of light, according to the relation \({n}_{1}=1.2+\frac{10.8 \times 10^{-14}}{\lambda^{2}} \text { and } {n}_{2}=1.45+\frac{1.8 \times 10^{-14}}{\lambda^{2}}\) The wavelength for which rays incident at any angle on the interface \(B C\) pass through without bending at that interface will be \(....\,nm.\)

- A \(500\)
- B \(600\)
- C \(700\)
- D \(800\)
Answer & Solution
Correct Answer
(B) \(600\)
Step-by-step Solution
Detailed explanation
For no deviation, \(n_{1}=n_{2}\) \(1.2+\frac{10.8 \times 10^{-14}}{\lambda^{2}} =1.45+\frac{1.8 \times 10^{-14}}{\lambda^{2}}\) \(0.25 =\frac{9 \times 10^{-14}}{\lambda^{2}}\) \(\lambda^{2} =\frac{9 \times 10^{-14}}{0.25}\) \(\lambda =\frac{3}{5} \times 10^{-6}\)…
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