A quarter cylinder of radius R, µ1 = 1.5 is placed on the table. A point object P is kept at a distance ‘mR’ from it, find the value of ’m’ for which a ray from ‘P’ will emerge parallel to the table. Consider all rays to be ……..
Solution: For first refraction, the object will appear to be kept at µmR from B for second refraction µ1 = 1 µ2 = 1
Diagram u = -(µmR + R) R = - R v = => => µm = 2 => m = s m = 4/3.
Medium
A cylinder with radius R = 0.1 m has µ = 3/2. Find the
(a) height ‘h’ of the ray so that the emergent ray emerges from , the end of a diameter parallel to surface.
(b) If the cylinder is kept over a plane mirror then at what distance from the centre of this cylinder rod, should identical rod be placed so that the final emergent ray is parallel to the incident ray.
Solution: Diagram Since ray is ll to AB BOC = i also BOC = 2r = i (exterior angle theorem) Now from Snell’s law sin i = µ sin r = sin r => sin 2r = sin r => 2 sin r cos r = sin r => cosr = 3/4 sin r = Now sin i = 2 sinr cosr = 2 x from CDO CD = h = CO sini = R sini =
(b) For the emergent ray to be parallel to the incident ray, the ray pattern must be symmetric.
Diagram in OAD tani = distance between centres = 2R + 2AD =
Equiconvex lens of refractive index 3/2 and f = 10 cm is held with its axis vertical and lower surface is immerged in water (µ = 4/3). The upper surface being in an. At what distance from the lens will a vertical beam of parallel light incident on lens to be focussed.
Solution: Diagram R = f = 10 cm For first refraction u = µ1 = 1 µ2 = 3/2 R = 10 cm v = 20 cm beam will be focussed at 20 cm from the lens inside the water.
A point ‘O’ is placed at a distance of 0.3 m from a convex lens, focal length 0.2 m, cut into2 halves each of which is displaced by 0.0005 m as shown in fig. Find the position of the image. If more than one image is formed. Find their numbers and distance between them.
Solution: Two images will be formed. Both pieces will act as single lens system with axis as the line joining object and pole of each segment.
Diagram for first lens object distance = OO1 = v = 0.6 = OA1 O1OD AOI
AI = 0.15 cm distance between two images = 2 AI = 2 x 0.15 = 0.3 m
A right angled prism (450 - 900 - 450) of refractive index n has a plate of reflective index n1 (< n) comented to its diagonal face. The assembly isin an a ray is incident on AB.
(a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.
(b) Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal undeviated.
Solution: Diagram in AOO’ 45 + (90 - r) + ( 90 - c) = 1800 r + c = 450 r = 450 - c sinr = (cosc - sinc) for critical angle C sinc = for refraction at first surface sin = n sinr =
(b) From Geometry it can be seen that r = 450, for ray to be incident normally on diagonal face.
Diagram taking refraction at first surafce. sin = n sin450
HARD
The x - y plane is boundary between two transparent media. Medium-1 with z 0 has a refractive index and medium 2 with z 0 has refractive index . A ray of light inmedium-1 given by vector is incident on the plane of separation, find the unit vector in the direction of the refracted ray in medium-2.
Solution: Let refracted ray br Normal to plane of incident and normal = =
Diagram it must also be normal to refracted ray cos ( - i) = = = cos1200s i = 600 r = 450 Now since angle between refracted ray and Normal = 450 c2 = a2 + b2 = a2 + c =
A ray of light in air is incident atgrazing angle (i = 900) on a long rectangular slab of a transparent medium of thickness t = 1.0 m. The point of incidence is the oprigin A(0, 0). The medium has a variable index of refraction n(y) given by n(y) = where k = 1.0 m-3/2. The refractive index of air is 1. (i) Obtain a relation between the slope of the trajectory of the ray at a point B(x, y) in the point. (ii) Obtain an equation for trajectory y(x) of the ray in the point. (iii) Determine the co-ordinates (x, y1) of the point P where the ray intensects the upper surface of the slab-air boundary. (d) Indicate the path of the ray subsequently.
Diagram
Solution: Diagram Taking on arbitrary point P(x, y) refractive index at this point n = from Snell’s law n sin = constant applying this for initail pt. (when ray is entering medium B) and at point. 1 x sin900 = sini sin i = it can be seen that i = s Slope = tan = cot i =
diagram
(ii) = coti = it passes throughorigin C = 0 x = 4 is the equation of trajectory when ray comes out of the mediums then x = 4 x 1 = 4 Co-ordiante of pt- is (4, 1) If medium on both sides are same, then angle with which the ray enters the medium = angle with which the ray comes out. ray will be parallel to x-axis
m = 4/3.
BOC = i also 
sin r
CDO
v = 20 cm
(cosc - sinc)
= n sinr = 
0 has a refractive index 
and medium 2 with z 
0 has refractive index 
. A ray of light inmedium-1 given by vector 
is incident on the plane of separation, find the unit vector in the direction of the refracted ray in medium-2.
- i) = 
= cos1200s
where k = 1.0 m-3/2. The refractive index of air is 1. (i) Obtain a relation between the slope of the trajectory of the ray at a point B(x, y) in the point. (ii) Obtain an equation for trajectory y(x) of the ray in the point. (iii) Determine the co-ordinates (x, y1) of the point P where the ray intensects the upper surface of the slab-air boundary. (d) Indicate the path of the ray subsequently.
sini
it can be seen that i = 
is the equation of trajectory