Optics Ques 38
- A monochromatic beam of light of wavelength $6000 $ $\AA$ in vacuum enters a medium of refractive index $1.5$. In the medium its wavelength is and its frequency is ……
$(1985,2 M)$
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Answer:
Correct Answer: 38.$4000 \AA, 5 \times 10^{14} Hz$
Solution:
- $\lambda _{\text {medium }}=\frac{\lambda _{\text {air }}}{\mu}=\frac{6000}{1.5}=4000 \AA$
$ f=\frac{v _{\text {air }}}{\lambda _{\text {air }}}=\frac{v _{\text {medium }}}{\lambda _{\text {medium }}}=\frac{3.0 \times 10^{8}}{6.0 \times 10^{-7}}=5.0 \times 10^{14} Hz $
Frequency remains unchanged.
BETA
