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An astronomical telescope is said to be in NORMAL adjustment when the distance between the objective lens and the eyepiece equals the sum of their focal lengths. hence, D = \(F_e\) + \(F_o\) = 50cm + 5cm = 55cm

The general equation for a wave is given by y = Asin⁡(k(x−vt)), where A is the amplitude, k is the wave number, v is the wave speed, and x and t are the position and time, respectively. The wave number k is related to the wavelength \lambda by the equation k = \(\frac{2\pi}{\lambda}\) In the given wave equation, y=0.5sin⁡(0.4\pi(x−60t)), the wave number k is 0.4\pi. Setting k = \(\frac{2\pi}{\lambda}\) equal to 0.4\pi and solving for \lambda gives: \lambda = \(\frac{2\pi}{0.4\pi}\) = 5cm So, the wavelength of the wave is 5 cm.

(a) A beam of polarised light is on constrained in a single plane perpendicular to the beam.

Table of values Icm\(^{3}\) hcm\(^{3}\) 10.0 7.50 15.0 11.25 20.0 15.00 25.0 18.75 30.0 22.50 Slope (s) = \(\frac{\bigtriangleup {h}}{\bigtriangleup {|}} = \frac{12.50}{20.00}\) = 0.625 Evaluate: (i) \(K_1\) = 1 - S = 1 - 0.625 = 0.375 (ii) \(K_2\) = \(\frac{1}{K} = \frac{1}{0.375}\) = 0.67 Precautions: The lower level of the meniscus of water is read at eye level. Parallax error is avoided in reading the scale to determine h and I Zero error of metre rule is noted. (b)0) Total internal reflection of light occurs when light travels from a dense medium to a less dense medium so that the angle of incidence (in the dense medium) is greater than the critical angle ii. Let apparent upward displacement of points be t, real depth, D = 6cm. Therefore apparent depth = D - t = 6 - t Refractive index aug =1.5 = \(\frac{\text {Real depth}}{\text {Apparent depth}}\) = 1.5 = \(\frac{6}{6 - t}\), (6-t) x 1.5 = 6cm \(\therefore\) 9-1.5t = 6 i.e t = \(\frac{9-6}{1.5}\) = 2.0cm OR Refractive index a\(\mu\)g = \(\frac{\text {real depth}}{\text {apparent depth}}\) \(\therefore\) Apparent depth = \(\frac{\text {real depth}}{a \mu g}\) = 6 = 4 1.5 Displacement = real depth - apparent depth = (6 - 4) cm = 2.0cm

The necessary conditions to produce interference fringes are I (Coherence), II (Same Frequency), and III (Same Wavelength). Condition IV (Same Intensity) is not necessary.

V = 72m U = ?, Magnification = 24 But Magnification = V/U ∴ 24 = 72/U U = 72/24 U = 3.0m

The vibrator is fixed at the end of the rope, hence, it is stationary. Also, the vibrator causes the end of the wire to move up and down (or side to side), creating disturbances that travel along the length of the wire. The displacement of the particles in the wire is perpendicular to the direction of wave propagation, which is characteristic of transverse waves Therefore, Option B is correct.

An astronomical telescope traditionally consists of two converging lenses with the objective having a longer focal length than the eyepiece. In this case, the 50 cm lens would serve as the objective, and the 5 cm lens as the eyepiece, which is a standard configuration for astronomical telescopes.

For a vibrating sonometer wire, the frequency F = l/2l \sqrt(T/m) and if the length, l and mass per unit length, m, are constant, then Fo ∝ \sqrtT => FoK\sqrtT or Fo/\sqrtT = K => F1o/\sqrtF1 = F2o/\sqrtF2 ∴ F1o/\sqrtT1 = F2o/\sqrtT2 thus F2 = F3x\sqrt(2T1 __\sqrtT1 = = Fox\sqrt2 x \sqrtT1 \sqrtT1 ∴F2 = Fo x \sqrt2 = Fox\sqrt2

The general wave equation is given by Y = sin A 2π/λ [x - vt] = Sin A [(2π X)/λ - (2π vt)/λ] thus if A = 0.15;λ = 0.01m; f = 550Hz then V = f x λ = 550 x 0.01 = 5.50m/s ∴ Y = 0.15 sin [(2 x λ x X)/0.01 - (2 x λ x 5.5t)/0.01] = 0.15 sin[200λX - 200 x λ x 5.5t] ∴ Y = 0.15 sin 200λ[x - 5.5t]

Angle of deviation = 180 - [i + v] = 180 - 2i [since i = v] ∴ 120 = 180 - 2i 2i = 180 - 120 = 60 ∴ i = 30o

As the light enters the dense medium, the shorter wavelengths (blue and violet) bend more than the longer wavelengths (red and orange), resulting in the separation of the light into its component colors. Based on this, the correct representation is: > Ray a represents the red color > Ray b represents the green color > Ray c represents the blue color

Given: distance(d) = 20.0m. time(t) = 1secs, f = 2 x 10\(^3\)Hz, wavelength(\lambda) = ? But v = \(\frac{s}{t}\) =20.0m/s = f\lambda \lambda = \(\frac{\text{v}}{\text{\lambda}}\) = \(\frac{20}{2\times10^3}\) = 0.01m = 1.0 x 10\(^{-2}\)m

y = A sin(\omegat + Φ ) A = 4cm, Φ = \(\frac{2}{3} \pi \) rad, \omega = \(\frac{1}{3}\pi\ rads^{-1}\) substituting the given values y = 4 sin ( \(\frac{1}{3}\pi\)t + \(\frac{2}{3} \pi \)) y = 4sin \(\frac{\pi}{3}\)( t + 2)

The reflected ray will rotate through angle 2 x 20º = 40º The reflected ray will now make angle 30º + 30º + 40º = 100º with the incident ray(30º + 30º since i = r)