PDF chapter test

To understand the concept of refraction of light through a glass slab, let us do an activity.
• Using drawing pins, fix a sheet of white paper to a drawing board.
• Cover the sheet in the middle with a rectangular glass slab.
• With a pencil, trace the slab's outline.
• Let us name the sketch as ABCD.
• Take four identical pins and arrange them in a row.
• Fix two pins vertically, say E and F, so that the line connecting them is inclined to the edge AB.
• Look through the opposite edge for the images of pins E and F.
• Fix two additional pins, say G and H, so that they, along with the images of E and F, form a straight line. Remove the slab and pins.
• Make a line up to AB by joining the positions of the tips of the pins E and F. Let EF meet AB at O.
• Similarly, join the positions of the tip of the pins G and H and produce it up to the edge CD.
• Let HG meet CD at O′
• Join O and O′.
• Also, produce EF up to P, as shown by a dotted line in the below diagram.
Refraction through a glass slab

You will notice that the light ray has changed direction at points $$O$$ and $$O′$$ in this activity. Both the points O and O′ are located on surfaces that separate two transparent media. At $$O′$$, draw a perpendicular NN' to AB and a perpendicular MM′ to CD. At point O, the light ray has crossed from a rarer medium to a denser medium, i.e., from air to glass. In this case, the light rays bend towards the normal. The light ray has crossed from glass to air at O′, moving from a denser to a rarer medium. The light ray moves away from the normal in this case. At both refracting surfaces AB and CD, compare the angle of incidence with the angle of refraction. The incident ray is a ray that is obliquely incident on the surface AB. The refracted ray is OO′, and the emergent ray is O′ H.

You will notice that the emergent ray follows the same path as the incident ray. At the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab, the extent of bending the ray of light is equal and opposite. The ray emerges parallel to the incident ray as a result of this. However, the light ray is slightly shifted to the side.

When a light ray is an incident normally on the interface of two media, what happens? Let us try to figure it out in the next section.