The mean and variance of five observations are 4.4 and 8.24 respectively.
Question: The mean and variance of five observations are 4.4 and 8.24 respectively. If three of these are 1, 2 and 6, find the other two observations. Solution: Given: Mean of 5 observations = 4.4 and Variance of 5 observations $=8.24$ Let the other two observations be $x$ and $y$ $\therefore$, our observations are $1,2,6, \mathrm{x}$ and $\mathrm{y}$ Now, we know that, Mean $(\overline{\mathrm{x}})=\frac{\text { Sum of observations }}{\text { Total number of observations }}$ $4.4=\frac{1+2+6+x+...
Read More →Evaluate the integral:
Question: Evaluate the integral: $\int \frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x$ Solution: $I=\int \frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x=\int \frac{(3 \sin x-2) \cos x}{5-\left(1-\sin ^{2} x\right)-4 \sin x} d x$ $\Rightarrow I=\int \frac{(3 \sin x-2) \cos x}{4+\sin ^{2} x-4 \sin x} d x$ Let, $\sin x=t \Rightarrow \cos x d x=d t$ $\therefore I=\int \frac{(3 t-2)}{t^{2}-4 t+4} d t$ As we can see that there is a term of $t$ in numerator and derivative of $t^{2}$ is a...
Read More →The mean and variance of five observations are 6 and 4 respectively.
Question: The mean and variance of five observations are 6 and 4 respectively. If three of these are 5, 7 and 9, find the other two observations. Solution: Given: Mean of 5 observations = 6 and Variance of 5 observations $=4$ Let the other two observations be $x$ and $y$ $\therefore$, our observations are $5,7,9, \mathrm{x}$ and $\mathrm{y}$ Now, we know that, $\operatorname{Mean}(\overline{\mathrm{x}})=\frac{\text { Sum of observations }}{\text { Total number of observations }}$ $6=\frac{5+7+9+...
Read More →(i) Consider a thin lens placed between
Question: (i) Consider a thin lens placed between a source (S) and an observer (O). Let the thickness of the lens vary as 2 0 () = b wb w, where b is the verticle distance from the pole. w0 is a constant. Using Fermats principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length. (ii) A gravitational lens may be assumed to have a varying...
Read More →An infinitely long cylinder of radius R is made
Question: An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index 1. The cylinder is placed between two planes whose normals are along the y-direction. The centre of the cylinder O lies along the y-axis. A narrow laser beam is directed along the y-direction from the lower plate. The laser source is at a horizontal distance x from the diameter in the y-direction. Find the range of x such that light emitted from the lower plane does not reach the upper p...
Read More →If light passes near a massive object,
Question: If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by n(r) = 1 + 2 GM/rc2 where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the r...
Read More →The mixture a pure liquid and a solution in a long vertical column
Question: The mixture a pure liquid and a solution in a long vertical column (i.e, horizontal dimensions vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical deviates from its original path. Find the deviation in travelling a horizontal distance d h, the height of the column. Solution: Let the height of the long vertical column with transparent liquid be h...
Read More →Show that for a material with refractive index µ ≥ 2 ,
Question: Show that for a material with refractive index 2 , light incident at any angle shall be guided along a length perpendicular to the incident face. Solution: Let the refractive index of the rectangular slab be 2. = 1/sin ic sin ic 1/ cos r 1/ sin i/sin r = From Snells law Sin I = sin r i = 90o 1 + 1 2 2 2 Taking the square root 2...
Read More →A myopic adult has a far point at 0.1 m.
Question: A myopic adult has a far point at 0.1 m. His power of accomodation is 4 diopters. (i) What power lenses are required to see distant objects? (ii) What is his near point without glasses? (iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.) Solution: (i) Power lenses are required to see distant objects 1/f = 1/v 1/u 1/f = 1/10 f = -10 cm = -0.1 m P = 1/f P = 1/(-0.1) P = -10 diopter (ii) When no corrective lens used Pn = ...
Read More →A jar of height h is filled with a transparent liquid
Question: A jar of height h is filled with a transparent liquid of refractive index . At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible. Solution: tan icd/2/h ic = d/2h d = 2h tan ic d = 2h 1/2 1...
Read More →In many experimental set-ups the source and screen
Question: In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points. Solution: u = -x1 v = +(D x1) 1/D x1 1/(-x1) = 1/f u = -x2 v = +(D x2) 1/D x2 1/(-x2) = 1/f D = x1 + x2 d = x2 x1 x1 = D d/2 D x1 = D + d/2 u = D/2 + d/2 v = D/2 d/2 m1 = D d/D + d m2/m1 = (D+d/D-d)2...
Read More →A thin convex lens of focal length 25 cm
Question: A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0,0) and an object placed at (50 cm, 0). Find the coordinates of the image. Solution: 1/v = 1/u + 1/f = 1/-50 + 1/25 = 1/50 v = 50 cm Magnification is m = v/u = -50/50 = -1 Therefore, the coordinates of the image are (50 cm, -1 cm)...
Read More →A circular disc of radius ‘R’ is placed co-axially and horizontally
Question: A circular disc of radius R is placed co-axially and horizontally inside an opaque hemispherical bowl of radius a. The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed? Solution: Distance at which the bowl should be placed in the disc is given as: d = (a2 b2)/(a + r)2 (a r)2...
Read More →The variance of 20 observations is 5. If each observation is multiplied by 2.
Question: The variance of 20 observations is 5. If each observation is multiplied by 2. Find the variance of the resulting observations Solution: Let the observations are $\mathrm{X}_{1}, \mathrm{X}_{2}, \mathrm{X}_{3}, \mathrm{X}_{4}, \ldots, \mathrm{X}_{20}$ and Let mean $=\overline{\mathrm{x}}$ Given: Variance = 5 and n = 20 We know that, Variance, $\sigma^{2}=\frac{1}{\mathrm{n}} \sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$ Putting the given values, we get $5=\frac{1}{...
Read More →A short object of length L is placed along
Question: A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take L |v-f| Solution: The mirror formula is 1/v + 1/u = 1/f u is the object distance v is the image distance du = |u1 u2| = L Differentiating on the both sides we get, dv/v2= -du/u2 v/u = f/u-f du = L, therefore, image length is f2/(u-f)2L...
Read More →For a glass prism (µ = √3 ) the angle of minimum
Question: For a glass prism ( = 3 ) the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism. Solution: = sin[(A + m)/2]/sin (A/2) Substituting the values of = sin A/sin A/2 = sin A/sin A/2 = 2 cos A/2 A = 60o Therefore, the required angle is 60o...
Read More →Three immiscible liquids of densities d1 > d2 > d3
Question: Three immiscible liquids of densities d1 d2 d3 and refractive indices 1 2 3 are put in a beaker. The height of each liquid column is h/3. A dot is made at the bottom of the beaker. For near-normal vision, find the apparent depth of the dot. Solution: h1 = -2/ 1 h/3 h2 = 3/ 2(2/ 1 h/3 + h/3) = -h/3 (3/ 2 + 2/ 1) h3(1/ 1 + 1/ 2 + 1/ 3) is the required depth....
Read More →An unsymmetrical double convex thin lens forms
Question: An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed? Solution: The position of the image will not change if the lens is reversed....
Read More →The near vision of an average person is 25cm.
Question: The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope? Solution: D = 25 cm u = -f v = -25 m = v/u f = 0.025 m p = 1/f = 40 D...
Read More →Will the focal length of a lens for red light be more,
Question: Will the focal length of a lens for red light be more, same or less than that for blue light? Solution: The focal length of a lens for red light will be larger than that for the blue light....
Read More →Evaluate the integral:
Question: Evaluate the integral: $\int \frac{a x^{3}+b x}{x^{4}+c^{2}} d x$ Solution: $I=\int \frac{a x^{2}+b x}{x^{4}+c^{2}} d x$ As we can see that there is a term of $x^{3}$ in numerator and derivative of $x^{4}$ is also $4 x^{3} .$ So there is a chance that we can make substitution for $x^{4}+c^{2}$ and I can be reduced to a fundamental integration but there is also a $x$ term present. So it is better to break this integration. $I=\int \frac{a x^{3}}{x^{4}+c^{2}} d x+\int \frac{b x}{x^{4}+c^...
Read More →An astronomical refractive telescope has an objective
Question: An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2cm. (a) The length of the telescope tube is 20.02m. (b) The magnification is 1000. (c) The image formed is inverted. (d) An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image. Solution: (a) The length of the telescope tube is 20.02m. (b) The magnification is 1000. (c) The image formed is inverted....
Read More →A magnifying glass is used,
Question: A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in (a) a larger angle to be subtended by the object at the eye and hence viewed in greater detail. (b) the formation of a virtual erect image. (c) increase in the field of view. (d) infinite magnification at the near point. Solution: (a) a larger angle to be subtended by the object at the eye and hence viewed in greater detail. (b) the formation of a virtual ...
Read More →Between the primary and secondary rainbows,
Question: Between the primary and secondary rainbows, there is a dark band known as Alexandars dark band. This is because (a) light scattered into this region interfere destructively. (b) there is no light scattered into this region (c) light is absorbed in this region. (d) angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42 and 50. Solution: (a) light scattered into this region interfere destructively. (d) angle made at the eye...
Read More →A rectangular block of glass ABCD has a refractive index
Question: A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB. When observed from the face AD, the pin shall A B DC (a) appear to be near A. (b) appear to be near D. (c) appear to be at the centre of AD. (d) not be seen at all. Solution: (a) appear to be near A. (d) not be seen at all....
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