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Chapter 16: Problem 2311

A sound source emits sound of \(600 \mathrm{~Hz}\) frequency, this sound enters by opened door of width \(0.75 \mathrm{~m} .\) Find the angle on one side at which first minimum is formed. The speed of sound \(=300 \mathrm{~ms}^{-1}\) (A) \(84.4^{\circ}\) (B) \(90^{\circ}\) (C) \(74.2^{\circ}\) (D) \(41.2^{\circ}\)

Short Answer

Expert verified
The angle at which the first minimum is formed when sound enters through the opened door is approximately \(41.8^{\circ}\). The correct answer is (D) \(41.2^{\circ}\).

Step by step solution

01

Calculate the wavelength

To find the wavelength of the sound, use the formula: \(\lambda = \frac{v}{f}\) Where: - \(v\) is the speed of sound (given as 300 m/s), and - \(f\) is the frequency of the sound (given as 600 Hz). Plugging in the values: \(\lambda = \frac{300 \mathrm{ms^{-1}}}{600 \mathrm{Hz}} = 0.5 \mathrm{m}\)
02

Use single-slit diffraction formula

Now, we can use the formula for the angle of the first minimum in single-slit diffraction: \(sin(\theta) = \frac{m\lambda}{a}\) Using the given values and the calculated wavelength: - \(m\) is the order of the minimum (1, for the first minimum), - \(\lambda\) is 0.5 meters, - \(a\) is the width of the door (0.75 meters). Plugging in the values: \(sin(\theta) = \frac{1 \cdot 0.5 \mathrm{m}}{0.75 \mathrm{m}} = \frac{2}{3}\)
03

Calculate the angle

Now we have the value for the sine of the angle, we can calculate the angle itself: \(\theta = sin^{-1}(\frac{2}{3})\) After calculating, we find: \(\theta \approx 41.81^{\circ}\) If we look at the given options, we see that (D) is the closest option to our calculated angle.
04

Conclusion

Therefore, the angle at which the first minimum is formed when sound enters through the opened door is approximately \(41.8^{\circ}\) . The correct answer is (D) \(41.2^{\circ}\).

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Most popular questions from this chapter

If the refractive index of a material of an equilateral Prism is \(\sqrt{3}\), then angle of minimum deviation will be (A) \(50^{\circ}\) (B) \(60^{\circ}\) (C) \(39^{\circ}\) (D) \(\overline{49^{\circ}}\)

One convex lens and one concave lens placed is contact with each other. If the ratio of their power is \((2 / 3)\) and focal length of the combination is $30 \mathrm{~cm}$, then individual focal lengths are (A) \(15 \mathrm{~cm}\) and \(-10 \mathrm{~cm}\) (B) \(-15 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) (C) \(30 \mathrm{~cm}\) and \(-20 \mathrm{~cm}\) (D) \(-30 \mathrm{~cm}\) and \(-30 \mathrm{~cm}\)

A Slit of width \(12 \times 10^{-7} \mathrm{~m}\) is illuminated by light of wavelength \(6000 \AA\). The angular width of the central maxima is approximately (A) \(30^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(0^{\circ}\)

\(1.6\) is a refractive index of plano-convex lens, then the radius of curvature of the curved surface is \(60 \mathrm{~cm}\). The focal length of the lens is \(\mathrm{cm}\) (B) \(10 \overline{0}\) (A) 50 (C) \(-50\) (D) \(-100\)

$$ \begin{array}{|l|l|} \hline \text { Column - I } & \text { Column - II } \\ \hline \text { (i) While going from rarer to denser medium } & \text { (a) Wavelength changes } \\ \text { (ii) While going from denser to rarer medium } & \text { (b) } \eta=(\mathrm{C} / \mathrm{V}) \\ \text { (iii) While going to one medium to another } & \text { (C) Ray bends towards normal } \\ \text { (iv) Refractive index of medium } & \text { (D) Rav bends awav from normal } \\ \hline \end{array} $$ (A) \(i-c\), ii \(-d\), iii \(-b\), iv-a (B) \(\mathrm{i}-\mathrm{a}\), ii \(-\mathrm{b}\), iii $-\mathrm{c}, \mathrm{iv}-\mathrm{d}$ (C) $\mathrm{i}-\mathrm{c}, \mathrm{ii}-\mathrm{b}, \mathrm{iii}-\mathrm{a}, \mathrm{iv}-\mathrm{d}$ (D) \(i-d, 1 i-c, 11 i-b, i v-a\)
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