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1. \(f_o\) = 15cm 5. a = 60.0cm, b = 20.0cm Hence, L = \(\frac{a}{b} = \frac{60}{20}\) = 3 6. Tables of values/observation S/N b(cm) a(cm) L=\(\frac{a}{b}\) 1 20.0 60.00 3.00 2 25.0 37.00 1.50 3 30.0 30.00 1.00 4 35.0 26.00 0.75 5 40.0 24.00 0.6 7. 8. Slope (s) = \(\frac{\bigtriangleup {L }}{\bigtriangleup {a}} = \frac{3-0.25}{60-18.6cm}\) = \(\frac{2.75}{41.4}\) = 0.066425cm\(^{-1}\) S\(^{-1}\) = \(\frac{1}{s} = \frac{1}{0.066425(cm)^{-1}}\) = 15.05 = 15 N.B: S\(^{-1}\) = \(f_o\) = 15cm Precautions taken to ensure accurate results are as follows: Parallax error in reading meter rule avoided Readings repeated must De shown on the table Collinear alignment of apparatus ensured (b)i. u = 10cm, f = +15cm, V= ? \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f }\) \(\frac{1}{v} + \frac{1}{10} = \frac{1}{15}\) \(\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} =\frac{1}{30}\) = The characteristics of the image formed are: - it is virtual - it is enlarged or magnified i.e twice or two times as big as the object (m = 2)

In vibration in a closed pipe/tube, the first resonance occurs at a position L = V/4f, When reading from the top of the tube, L\(_1\) = \(\frac{340}{4 \times 340}\) = 0.25m = 0.25m L\(_2\) = \(\frac{3V}{4f}\) L\(_2\) = \(\frac{3 \times 340}{4 \times 340}\) = 0.75m L\(_3\) = \(\frac{5V}{4f}\) L\(_3\) = \(\frac{5\times 340}{4\times 340}\) = 1.25m However, the effective length of the tube is 1.2m. Thus reading from below, the first position of resonance will be 1.20 - 0.25m = 0.95m

Form the lens formula = \(\frac{1}{\text{U}}\) + \(\frac{1}{\text{V}}\) = \(\frac{1}{\text{f}}\) => \(\frac{1}{10} + \frac{1}{25}\) = \(\frac{1}{\text{f}}\) ∴ \(\frac{(5 + 2)}{50}\) = \(\frac{1}{\text{f}}\) ∴ f = \(\frac{50}{7}\)cm Since the image is observed on a screen, it is a real image. Thus magnification M =\(\frac{\text{f}}{U - f}\) = \(\frac{\frac{50}{7}}{10 - \frac{50}{7}}\) ∴ M = \(\frac{\frac{50}{7}}{\frac{20}{7}}\) = \(\frac{50}{7}\) x \(\frac{7}{20}\) = 2.5 Or m = \(\frac{\text{v}}{\text{u}}\) = \(\frac{25}{10}\) = 2.5

The number of images to be observed will be infinite

From the above diagram the widest separation is given by Blue and red

At minimum deviation, it is observed that light passes symmetrically through, the prism such that the angle of incidence (io) is angle to the angle of emergency (eo).(i/e Io = eo)

For a concave mirror to produce an image at infinity, the image has to be place at the principal focus of the mirror

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

Any image that can be formed on a screen are called real image. Such images are formed by light rays emanating from the object, when reflected and/or refracted by devices such as lenses and mirrors converge at some specific point. In contrast there are other images which are formed by the rays of light reaching the observer appearing to converge at some point. This happens when the rays reaching the observer are diverging and when the line of these rays are projected backward. Such images, which cannot be formed on a screen, are called virtual image.

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

M = \(\frac{v}{u}\) v = mu = 4 x 25 = 100m \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) \(\frac{1}{100} + \frac{1}{25} = \frac{1}{f}\) f = 20cm

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

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 C.

For the equilateral glass prism, A = 60º n = \(\frac{sin \frac{Dm + A}{2}}{sin \frac{A}{2}}\) \(\frac{3}{2}\) = \(\frac{sin \frac{Dm + 60}{2}}{sin \frac{60}{2}}\) = \(\frac{sin \frac{Dm + 60}{2}}{sin 30}\) \(\frac{3}{2}\) x \(\frac{1}{2}\) = sin\(\frac{Dm + 60}{2}\) \(\frac{3}{4}\) = 0.75 = sin\(\frac{Dm + 60}{2}\) sin\(^{-1}\)(0.75) = \(\frac{Dm + 60}{2}\) 48.590 = \(\frac{Dm + 60}{2}\) 48.590 x 2 = Dm + 60 97.180 = Dm + 60 Dm = 97.180 - 60 = 37.18 ≈ 37 .2º