Your cart is empty.
Practice each question, then open the answer or worked solution.
For a glass prism (µ = √3 ), the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.
μ = sin[(A + δm)/2]/sin (A/2)
Substituting the values of μ = sin A/sin A/2
μ = sin A/sin A/2 = 2 cos A/2
A = 60o
Therefore, the required angle is 60o
Short AnswersA short object of length L is placed along the principal axis of a concave mirror away from the 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|
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 both sides, we get,
dv/v2 = -du/u2
v/u = f/u-f
du = L, therefore, image length is f2/(u-f)2L
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 µ, and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?
The distance at which the bowl should be placed in the disc is given as:
d = μ(a2 – b2)/√(a + r)2 – μ(a – r)2
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 is placed at (–50 cm, 0). Find the coordinates of the image.
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)
In many experimental set-ups, the source and screen are fixed at a distance, say D and the lens are 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.
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
A jar of height h is filled with a transparent liquid of refractive index µ. 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.
tan ic d/2/h
ic = d/2h
d = 2h tan ic
d = 2h ×1/√μ2 – 1
A myopic adult has a far point at 0.1 m. His power of accommodation 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.)
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 is used
Pn = Pf + Pa
u = -10 cm = -0.1 m
v = 2 cm = 0.02 m
1/f = 1/v – 1/u = 60
Pn = Pf + Pa = 64 D
u = xn
xn = 1/14 m = 100/14 cm = 7 cm
iii) When a corrective lens is used
u = infinity
v = 2 cm = 0.02 m
1/f = 1/v – 1/u
P = 50
Pn = 50 + 0
Pn = 50 D
u = xnm
xm = ¼ m = 0.25 m
Show that for a material with refractive index µ ≥ 2, light incident at any angle shall be guided along a length perpendicular to the incident face.
Let the refractive index of the rectangular slab be μ ≥ √2.
μ = 1/sin ic
sin ic > 1/ μ
cos r ≥ 1/ μ
sin i/sin r = μ
From Snell’s law
Sin I = μ sin r
i = 90o
1 + 1 ≤ μ2
2 ≤ μ2
Taking the square root
√2 ≤ μ
The mixture of 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.
Let the height of the long vertical column with transparent liquid be h and dx be the thickness
The angle at which the ray AB enters is θ
Let y be the new height of the liquid
(θ + d θ) is the emerging angle while (y + dy) is the height.
From Snell’s law,
μ(y) sin θ = μ (y + dy) sin (θ + d θ)
After integration, we get,
θ = -1/ μ d μ/dy d
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 ray from the original path as it grazes the object.
n(r) = 1 + 2GM/rc2