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REFRACTION THROUGH GLASS SLAB (lateral and vertical shifts)

REFRACTION THROUGH GLASS SLAB: A thin glass slab is formed when a medium is isolated from its surroundings by two plane surfaces parallel to each other. Let us determine position and nature of the image formed when the slab is placed in front of an object. Lateral shift: Understanding with an Activity: Material required: plank, chart paper, clamps, scale, pencil, thin glass slab and pins. Procedure: Place a piece of chart (paper) on a plank. Clamp it. Place a glass slab in the middle of the paper. Draw border line along the edges of the slab by using a pencil. Remove it. You will get a figure of a rectangle. Name the vertices of the rectangle as A, B, C and D. Draw a perpendicular at a point on the longer side (AB) of the rectangle. Again keep the slab on paper such that it coincides with the sides of the rectangle ABCD. Take two pins. Stick them on the perpendicular line to AB. Take two more pins and stick them on the other side of the slab in such a ...

Total internal reflection (T.I.R)

TOTAL INTERNAL REFLECTION: We know that when light enters from denser medium to rarer medium it bends away to the normal.(i<r) What happens when angle of incidence goes on increasing?. Observe the diagram below. So as we increase angle of incidence the refracted ray continue to bend further away from normal and at one particular angle of incidence it grazes through the interface. This angle is known as critical angle. CRITICAL ANGLE: The angle of incidence for which refracted ray grazes through the interface when light is sent from denser to rarer medium is called critical angle. At critical angle, i=c and r=90 0 Let C be the critical angle. Then r becomes 90 0 we get, µ1/µ2 = sin 90/ sin c (applying Snell’s law) => µ 1 /µ 2 = 1/sin c . We get sin c = µ 2 / µ 1 .We know that µ 1 /µ 2 i.e., µ 12 is called refractive index of denser medium with respect to rarer medium sin c = 1/µ 12 What happens when angle of incidence is...

Example questions from refraction through curved surfaces

Example questions 1 1) Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged? Explain with a neat ray diagram. ANS: i) A parallel beam of light rays will converge on focal point of the lens after refraction. ii) Light rays passes through the focal point will parallel to principal axis after re-fraction. iii) So the two lenses are arranged on a common principal axis such that their focal points coincide with each other, then the rays remain parallel after passing through both lenses. 2 2) The focal length of a converging lens is 20cm. An object is 60cm from the lens. Where will the image be formed and what kind of image is it? ANS : Given, f=20 cm U=-60cm ...

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