Question

Monochromatic light of wavelength 589 mm is incident from air on a water surface .What are the wavelengths, frequency and speed of (a) reflected and (b) refracted rays? Refractive index of water is 1.33.

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Answer 1

Answer:

Hence the speed frequency and wavelength of the refliegth of the reflected light are 3×108 m/s 5×1014 Hz and 589 nm respecticely . Hence th speed freuency and wavelength of refracted light are 2.26×108 m/s 444.01 nm and 5.09×1014 Hz respectively .

Explanation:

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

Monochromatic light of wavelength 589 mm is incident from air on a water surface .What are the wavelengths, frequency and speed of :

(a) reflected ,and (b) refracted rays?

• Refractive index of water is 1.33.

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

Wavelength of  Monochromatic light , λ  = 589 nm

Speed of Light , c = 3 × 10⁸ m/s

Refractive index , μ = 1.33

(a) For reflected light ,

Wavelength is going to remain same for incident and reflected light.

Wavelength , λ = 589 nm = 589 × 10 ⁻⁹ m

Frequency of light , v

= c / λ

$\sf{=\dfrac{3 \times 10^{8}}{589 \times 10^{-9}}}$

= 5.09 × 10¹⁴ Hz

Speed of light will remain invariant in two media

Therefore, v = c = 3 × 10⁸ m/s

(b) For refracted light,

Refractive index of water with respect to air,

$\huge\boxed{\sf{\mathrm{a} \mu_{\omega}=\dfrac{\mathrm{v}_{\mathrm{a}}}{\mathrm{v}_{\omega}}}}$

i.e., $\huge\boxed{\sf{\mathrm{a} \mu_{\omega}=\dfrac{\mathrm{v}_{\mathrm{a}} \lambda_{\mathrm{a}}}{\mathrm{v}_{\omega} \lambda_{\omega}}}}$

∴ $\sf{v_a=v_\omega}$

So wavelength of light,

$\sf{\displaystyle\lambda_{\omega}=\frac{\lambda_{\mathrm{a}}}{\mathrm{a} \mu_{\omega}}=\frac{589 \times 10^{-9}}{1.33}}$

= 4.42 × 10⁻⁷ m

As frequency remains unaffected on entering another medium, therefore,

$\sf{v_{\omega}=v_{a}=5.09 \times 10^{14} \mathrm{\:Hz}}$

Speed,

$\sf{\displaystyle\mathrm{v}^{\prime}=\frac{V_{\mathrm{a}}}{\mu_{\omega}}=\frac{3 \times 10^{8}}{1.33}}$

= 2.25 × 10⁸ m/s

i.e., speed of the light has decresed when it entered the second medium.