The angle of minimum deviation for a prism depends on:
A. Refractive index only
B. Both angle of prism and refractive index
C. Angle of prism only
D. Wavelength of light only
Answer: Option B
Solution (By JKSSB Mock Tests)
Minimum deviation δ_m for a prism: n = sin[(A + δ_m)/2] / sin(A/2), where A is prism angle, n refractive index. Thus δ_m depends on both A and n. Since n varies with wavelength (dispersion), δ_m also depends on wavelength, but the primary dependencies are A and n. Memory tip: 'δ_m increases with A and n; used to measure n'. This optics formula application is frequently tested in competitive exams. Always recall that minimum deviation occurs when ray passes symmetrically through prism, a key condition for derivation.
Explanation:
Gamma rays have wavelengths < 10⁻¹² m, shortest in EM spectrum. X-rays 10⁻¹⁰-10⁻⁸ m. UV 10⁻⁸-4×10⁻⁷ m. IR longer. So gamma rays shortest. Highest frequency, energy.
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