(a) Define dffraction.
(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \(\alpha\) Determine the value of the critical angle. \(\beta\) Calculate the refractive index of the block.
(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:
(i) fundamental frequency of the open pipe;
(ii) length of the closed pipe;
(iii) length of the open pipe. [Speed of sound in air = 330 ms\(^{-1}\)]
(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \(\alpha\) Determine the value of the critical angle. \(\beta\) Calculate the refractive index of the block.
(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:
(i) fundamental frequency of the open pipe;
(ii) length of the closed pipe;
(iii) length of the open pipe. [Speed of sound in air = 330 ms\(^{-1}\)]
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) Diffraction is the bending or spreading of a wave round an obstacle or as it passes through an opening.
(b) (i) Critical angle is the angle of incidence in the denser medium when an angle of refraction in the lens dense medium is 90°.
(ii) (\(\alpha\))Critical angle = 90° -.44° = 46°
(\(\beta\))n = \(\frac{1}{sin c} = \frac{1}{\sin 46^o}\) = 1.39
(c) (i) Closed Pipe
1st Overtone = 3f\(_o\) = 3 x 200 = 600Hz
Open Pipe
All harmonics are possible overtone = 2f\(_o\)
2f\(_o\) = 600
f\(_o\) = 300Hz
(ii) Closed Pipe
f\(_o\) = \(\frac{v}{4l}\)
l = \(\frac{v}{4f_o}\)
\(\frac{330}{4 \times 200}\)
= 0.412m
Open Pipe
\(f_1\) = \(\frac{v}{l}\)
600 = \(\frac{300}{l}\)
l = 0.55m
or
\(f_1\) = \(\frac{v}{2l}\)
300 = \(\frac{300}{2 \times l}\)
l = \(\frac{330}{600}\)
= 0.55m
(a) Diffraction is the bending or spreading of a wave round an obstacle or as it passes through an opening.
(b) (i) Critical angle is the angle of incidence in the denser medium when an angle of refraction in the lens dense medium is 90°.
(ii) (\(\alpha\))Critical angle = 90° -.44° = 46°
(\(\beta\))n = \(\frac{1}{sin c} = \frac{1}{\sin 46^o}\) = 1.39
(c) (i) Closed Pipe
1st Overtone = 3f\(_o\) = 3 x 200 = 600Hz
Open Pipe
All harmonics are possible overtone = 2f\(_o\)
2f\(_o\) = 600
f\(_o\) = 300Hz
(ii) Closed Pipe
f\(_o\) = \(\frac{v}{4l}\)
l = \(\frac{v}{4f_o}\)
\(\frac{330}{4 \times 200}\)
= 0.412m
Open Pipe
\(f_1\) = \(\frac{v}{l}\)
600 = \(\frac{300}{l}\)
l = 0.55m
or
\(f_1\) = \(\frac{v}{2l}\)
300 = \(\frac{300}{2 \times l}\)
l = \(\frac{330}{600}\)
= 0.55m