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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.

The magnitude of the electric field strength at any point between the plates is not directly proportional to the magnitude of the charge on each plate. Instead, it is directly proportional to the charge density (charge per unit area) on the plates. Therefore, if the charge on the plates is increased while keeping the area of the plates constant, the electric field strength will increase. However, if the area of the plates is also increased such that the charge density remains constant, the electric field strength will not change.

Number of cycles, n = \(2\frac{1}{2} = \frac{5}{2}\) Length = n\lambda \lambda = \(\frac{Length}{n}\) = \(\frac{125}{\frac{5}{2}}\) Therefore, \lambda = 50cm = 0.50m.

F = qvB ( force on moving charge in a magnetic field) F = \(\frac{mv^2}{r}\) equating F qvB = \(\frac{mv^2}{r}\) qB = \(\frac{mv}{r}\) qBr = mv Therefore, v = \(\frac{qBr}{m}\)