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(a)(i) The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane. OR The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant \(\frac{\text{Sin i}}{\text{Sin r}}\) = Constant Place the triangular prism on the drawing sheet and trace the outline. Arrange two pins on one side vertically and using two other pins, determine the emergent ray on the other side. Remove the prism and. join the position of pins to meet the outline of the prism. Draw the normal at the point of incidence. Produce both the incident and the emergent rays to intersect and measure the angles of incidence I^\circ and the deviation d^\circ. Repeat the procedure for at least four other values of i. Plot a graph of d on the vertical axis against i on the horizontal axis. Obtain minimum deviation D as below. n = \(\frac{sin \frac{1}{2}(A +D)}{sin \frac{1}{2} A}\) = \(\frac{sin \frac{1}{2} (60 + D)}{sin 30}\) Precautions - neat traces - pins fixed vertically - Avoid parallax in reading protector (b)(i) (ii) n = \(\frac{\text{real depth}}{\text{apparent depth}}\) apparent depth = \(\frac{\text{real depth}}{n}\) = \(\frac{12}{1.5}\) = 8cm Apparent displacement = real depth - apparent depth = 12 - 8 = 4cm

Table os values/observation D(cm) D\(^{2}\)cm \(X_1\)(cm) \(X_2\)(cm) L=(\(x_1\)-\(x_2\))cm L\(^{2}\)cm D\(^{2}\)-L\(^{2}\)cm 100 10000 80.50 18.30 62.20 3868.8 6131.16 90 8100 70.20 19.00 51.20 2621.4 5478.56 80 6400 59.10 20.00 39.10 1528.8 4871.19 70 4900 46.70 22.00 24.70 610.09 4289.91 60 3600 29.00 7.60 21.40 457.96 3142.04 Slope (s) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{6000-3000}{100-50} = \frac{300}{50}\) = 60 Evaluate K = \(\frac{8}{4} = \frac{60}{4}\) = 15 Precaution: I avoided parallax when reading the metre rule. I ensured that a sharp image is formed on the Screen. (b) Real image is formed by the actual intersection of rays whereas a virtual image is formed by the apparent intersection of rays when their directions have been produced backwards, or a Real image is formed at the front of the mirror while a virtual image is formed at the back of the mirror. ii An object beyond 2F, the image is real, inverted, smaller than an object, between F and 2F, ray parallel to the principal axis and ray passes through the optical centre.

The lowest note (fundamental note) emitted by any stretched string has a fundamental frequecy fo = V/2l = 40Hz The 1st overtone = 2fo = 2 x 40 = 80Hz 2nd overtone = 3fo = 3 x 40 = 120Hz 3rd overtone = 4fo = 4 x 40 = 160Hz 4th overtone = 5fo = 5 x 40 = 200Hz Overtones between 40Hz and 180Hz are 80Hz, 120Hz, 160Hz = 3 Overtones

The image of an object placed in front of a plane mirror is always as far behind the mirror as the object is in front of the mirror. Thus when the object is 4m in front of the mirror, the image will be 4m behind the mirror. But when the mirror is moved in towards the man, (that is the man is now 3m from the mirror), the image distance will again decrease by 1m from behind the mirror. Thus the new image position is 3m behind the mirror. Therefore, the distance between the man and his new image is 6m.

Since diameter of the sphere = 20cm, then, the radius of curvature = 10cm Focal length, f = \(\frac{R}{2} = \frac{10}{2}\) = 5cm The inside of a hollow metal sphere forms a concave mirror.

Refractive index = \(\frac{Sin i}{Sin r}\) = \(\frac{\text{Vel. of light in air}}{\text{vel. of light in glass}}\) = \(\frac{Sin i}{Sin 30^o}\)= \(\frac{3.0 \times 10^8}{2.0 \times 10^8}\) = Sin i = \(\frac{(3.0 \times 10^8 \times Sin 30^o)}{2.0 \times 10^8}\) = Sin i = \(\frac{(3.0 \times 10^8 \times 0.5)}{2.0 \times 10^8}\) = \(\frac{1.5}{2}\) = 0.75

No expanded working has been attached for this question yet. The verified answer is Option A.

Inductive reactance XL = 2πfL ∴ 2πfL = 2000Ω then f = 2000Ω/2 x π x 10 f = 100/πHz

At resonance, inductive reactant XL is equal to the capacitive reactance XC i.e. XL = XC and XL = VL/IL and XC = VC/IC Again since the components are in series, the same current flows in the series arrangment i.e. IL = IC at resonance, VL/IL = VC/IC ∴ VL = VC = 40V

No expanded working has been attached for this question yet. The verified answer is Option C.

Given: - Frequency \( f = 12 \, \text{Hz} \) The period \( T \) is the reciprocal of the frequency: \(T = \frac{1}{f}\) Substituting the given frequency: \(T = \frac{1}{12 \, \text{Hz}} \approx 0.0833 \, \text{s}\) Thus, the period of the motion is approximately: \(T \approx 0.08 \, \text{s}\).

GIVEN: The angle of incidence i = 20º, Rotation of the mirror = 2 x 20º = 40º, Then the new angle of incidence i after rotation of mirror = 20º + 40º But, angle of reflection r = angle of incidence i ( law of reflection) Therefore, the angle between the incident ray and the reflected ray = i + r = 60º + 60º = 120º

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option C.