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As light ray enters a drop of water the light is refracted at the surface and at the end of the drop, it is totally internally reflected in which the reflected light returns to the front surface, where it again undergoes refraction as it moves from water to air. The result of this is a dispersed light of colours of different wavelengths.

The pinhole camera works on the principle of the rectilinear propagation of light. This principle states that light travels in straight lines. When light passes through the tiny hole in a pinhole camera, it forms an inverted image on the opposite side of the camera. The size of the image depends on the distance between the object and the pinhole.

f = 15cm (diverging mirror); u = -35cm(because object is in front of the mirror) v=? ⇒\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\) ⇒ \(\frac{1}{15}\) = \(\frac{1}{v}\) - \(\frac{1}{35}\) ⇒ \(\frac{1}{15}\) + \(\frac{1}{35}\) = \(\frac{1}{v}\) ⇒ \(\frac{105}{10}\) = \(\frac{1}{v}\) v = \(\frac{105}{10}\) = 10.5cm ∴The image is 10.5cm behind the mirror, since v is positive.

n=1.458, c=\(3.00 \times10^8 ms^-1\) ,\lambdao = 589nm; f=? Speed of light in a medium (v)=\(\frac{c}{n}\) where n is the refractive index of the medium ⇒\lambdan=\(\frac{589}{1.458}\) = 404nm v=f\lambda ⇒f=\(\frac{v}{\lambda}\) =\(\frac{2.06\times10^8}{404\times10^-9}\) \(1nano=10^{-9}\) ∴f=\(5.09\times10^{14}\) Hz

\(n_1\)=1.33; \(n_2\)=1.5; i=40^\circ; r= ? from snell's law: \(n_1\) x sin i=\(n_2\)x sin r ⇒1.33 x sin 40^\circ=1.5 x sin r ⇒sin r = 0.5700 ⇒r = \(sin^{-1}\) (0.5700) ⇒ r = 34.75

A=0.8m; f=16Hz; v=20 \(ms^-1\); k=? v = f\lambda \lambda =1.25m = k= \(\frac{2\pi}{\lambda}\) =\(\frac{2 \times 3.142}{1.25}\) k= 5m ( to 1 s.f)

Infrared waves are used in night vision goggles because they can penetrate the darkness and allow people to see objects that are not visible to the naked eye

Sound waves are mechanical waves that are longitudinal. This means that the particles of the medium through which sound waves travel, such as air or water, move back and forth in the same direction as the wave travels.

d=312m.,T =263\(-10^^\circc\) = -10+273K = 263K, t=? v= 331\sqrt \(\frac{T}{273}\) ⇒ v = 331\times\sqrt \(\frac{263}{273}\) ⇒ v = 324.88 \(ms^-1\) ⇒ v = \(\frac{d}{t}\) t = \(\frac{d}{t}\) = \(\frac {312}{324.88}\) t = 0.96s

The frequency of a sound wave is proportional to the tension of the string. If two instruments have the same frequency, but one has a tighter string, then the instrument with the tighter string will have a higher pitch. The other factors listed, such as the size of the instrument, the material of the instrument, and the shape of the instrument, will also affect the pitch of the instrument, but they will have a smaller effect than the tension of the string.

T=800N; I=50cm=0.5m, m=10g=0.01kg fundamental freq: \(f_o\) =? \(f_o\) = \(\frac{1}{21}\sqrt{T}{\mu}\) \mu =\(\frac{m}{1}\)=\(\frac{0.01}{0.5}\) ⇒ \(f_o\) =\(\frac{1}{2\times0.5}\)\sqrt\(\frac{800}{0.02}\) \(f_o\) ⇒\sqrt 40,000 ⇒1st overtone = 2\(f_o\) =2\times200 = 400Hz ⇒2nd overtone =3\(f_o\) =3\times200=600Hz ∴3rd over tone= 4\(f_o\) =4\times200=800Hz

Instrument resolution is not a limitation of experimental measurements. It is the smallest change in a measured quantity that can be detected by an instrument. While instrument resolution limits the accuracy of a measurement, it is not a limitation of experimental measurements itself.

Diffraction is the bending of waves around the corners of an obstacle or through an aperture.

When a negatively charged rod is brought close to a metal sphere, the free electrons in the sphere are repelled from the rod and move to the other end of the sphere. This creates a region of positive charge on the side of the sphere closest to the rod, and a region of negative charge on the opposite side. The process of charge distribution stops when the net force on the free electrons inside the metal is equal to zero. If the sphere is then earthed, the free electrons will flow from the sphere to the ground, leaving the sphere with a net positive charge.

When the temperature is decreased in a reversible reaction, the equilibrium shifts in the direction that produces heat (exothermic direction). For an exothermic reaction, this means the equilibrium position shifts to the right, favoring the forward reaction.