A taxi

For a certain optical medium the speed of light varies from a low value of 1.90 × 10 8 m/s for violet light to a high value of 2

Dmitry_Shevchenko [17]

Answer:

a. The refractive index ranges from 1.5 - 1.56

b. 18.7° for violet light and 19.5° for red light.

c. 33.7° for violet light and 35.3° for red light.

Explanation:

a. The refractive index of an object is the ratio of the speed of light in a vacuum and the speed of light in the object.

Mathematically,

$n = \frac{c}{v}$

The speed of violet light in the object is $1.9 * 10^8 m/s$ .

The speed of red light in the object is $2 * 10^8 m/s$

Hence, the refractive index for violet light is:

$n = \frac{3 * 10^8 }{1.9 * 10^8} \\\\n = 1.56$

and for red light, it is:

$n = \frac{3 * 10^8 }{2 * 10^8} \\\\n = 1.5$

Hence, the refractive index ranges from 1.5 - 1.56.

b. The refractive index is also the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

$n = \frac{sin(i)}{sin(r)}$

The angle of incidence is 30°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.56} = \frac{0.5}{1.56} \\\\sin(r) = 0.3205\\\\r = 18.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.5} = \frac{0.5}{1.5} \\\\sin(r) = 0.3333\\\\r = 19.5^o$

The angle of refraction for red light is larger than that of violet light when the angle of incidence is 30°.

c. The angle of incidence is 60°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.56} = \frac{0.8660}{1.56} \\\\sin(r) = 0.5551\\\\r = 33.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.5} = \frac{0.8660}{1.5} \\\\sin(r) = 0.5773\\\\r = 35.3^o$

The angle of refraction for red light is still larger than that of violet light when the angle of incidence is 60°.

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Explanation:

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Pulling the two ends of a rubber band further and further apart from each other.

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Church organs have a set of pipes with different lengths. With those different pipes organs can produce sounds over a wide range

Paraphin [41]

Answer: The shortest possible wavelength of sound the organ can produce is 0.224 m

Explanation:

To calculate the wavelength of light, we use the equation:

$\lambda=\frac{c}{\nu}$

where,

$\lambda$ = wavelength of the light

c = speed of light = $331m/s$

As wavelength and frequency follows inverse relation, shortest wavelength is produced by highest frequency.

$\nu$ = highest frequency of light = $1.48kHz=1.48\times 10^3Hz=1.48\times 10^3s^{-1}$ $1Hz=1s^{-1}$

$\lambda=\frac{331m/s}{1.48\times 10^3s^{-1}}=0.224m$

Thus the shortest possible wavelength of sound the organ can produce is 0.224 m

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