Question

1. a progressive wave of frequency 20 hz and speed 0.60 m s⁻¹ is reflected on itself from a rigid boundary such that the distance between a node and an adjacent antinode is x. determine the value of x.

a. 0.1200 m
b. 0.0600 m
c. 0.0150 m
d. 0.0075 m

1. when light from a source is propagated through a gas

i. certain wavelengths are absorbed.
ii. an emission line spectrum results.
iii. dark lines are observed.
which of the statements above are correct?
a. i and ii only
b. i and iii only
c. ii and iii only
d. i, ii and iii

1. a concave mirror of radius of curvature 20 cm forms an erect image 15 cm from the mirror. calculate the distance of the object from the mirror.

a. 60.0 cm
b. 30.0 cm
c. 8.5 cm
d. 6.0 cm

1. diagram: light ray at boundary, air and glass. rays a (incident), b (reflected), c (refracted?), d (maybe?)

in the diagram above, the refractive index of glass relative to air is
a. \\(\frac{\sin a}{\sin c}\\)
b. \\(\frac{\sin a}{\sin d}\\)
c. \\(\frac{\sin b}{\sin d}\\)
d. \\(\frac{\sin c}{\sin a}\\)

1. total internal reflection of light occurs in a medium when the

a. angle of incidence for a ray in the denser medium is greater than the critical angle.
b. angle of incidence for a ray in the denser medium is equal to the critical angle.
c. angle of incidence for a ray in the denser medium is less than the critical angle.

Explanation:

Question 25

Step1: Find wavelength using \( v = f\lambda \)

Given \( f = 20\space Hz \), \( v = 0.60\space m/s \). Rearranging \( \lambda=\frac{v}{f} \).
\( \lambda=\frac{0.60}{20}=0.03\space m \).

Step2: Distance between node and antinode

In a standing wave, distance between node and adjacent antinode is \( \frac{\lambda}{4} \).
\( x = \frac{0.03}{4}=0.0075\space m \).

• I: When light passes through gas, certain wavelengths (corresponding to energy transitions) are absorbed. Correct.
• II: Absorption (not emission) line spectrum is formed with dark lines. So II is incorrect.
• III: Absorbed wavelengths appear as dark lines in the spectrum. Correct.

So I and III are correct.

Step1: Recall mirror formula and sign conventions

For concave mirror, radius \( R = 20\space cm \), so focal length \( f=\frac{R}{2}=-10\space cm \) (negative as per sign convention, concave mirror). Image is erect, so \( v = +15\space cm \) (erect image in concave mirror is virtual, so \( v \) positive).

Mirror formula: \( \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \).

Step2: Solve for \( u \)

Substitute \( f = -10 \), \( v = 15 \):
\( \frac{1}{-10}=\frac{1}{u}+\frac{1}{15} \)
\( \frac{1}{u}=\frac{1}{-10}-\frac{1}{15}=\frac{-3 - 2}{30}=\frac{-5}{30}=-\frac{1}{6} \)
\( u = -6.0\space cm \)? Wait, no—wait, erect image in concave mirror: wait, concave mirror forms erect image only when object is within focal length, virtual image. Wait, maybe sign convention error. Wait, let's recheck:

Wait, concave mirror: \( f \) is negative (if using Cartesian sign convention, where incident light is from left, mirror at origin, \( f = -R/2 \)). Image is erect, so \( v \) is positive (virtual image, behind mirror).

So \( \frac{1}{u}=\frac{1}{f}-\frac{1}{v}=\frac{1}{-10}-\frac{1}{15}=\frac{-3 - 2}{30}=-\frac{5}{30}=-\frac{1}{6} \), so \( u = -6\space cm \)? But options have 6.0 cm. Wait, maybe sign convention where distances are measured as magnitudes with sign: for concave mirror, \( f = -10 \), \( v = +15 \) (virtual image). Then \( \frac{1}{u}=\frac{1}{f}-\frac{1}{v}=\frac{1}{-10}-\frac{1}{15}=-\frac{1}{6} \), so \( u = -6\space cm \), meaning object is 6 cm in front of mirror (since \( u \) is negative, real object). So distance is 6.0 cm.

Answer:

D. 0.0075 m

Question 26