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 on 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 form a straight line along with the images of $$E$$ and $$F$$. 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.

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