To find the refractive index of glass by real and apparant width of a glass slab.
APPARATUS:
Glass slab, drawing board, drawing pins, sheets of paper, cello tape and a pencil.
PRINCIPLE:
An object placed in a denser medium when viewed from a rarer medium will appear to be at a lesser distance because of refraction.
FORMULA:
Students are advised to leave four lines here
PROCEDURE:
1. Fix a sheet of paper on the drawing board with cello tape.
2. Place the glass slab horizontally and draw its boundary PQRS.
3. Remove the glass slab.
4. Mark a point O on PQ as shown in the figure and draw the normal NON’ through O.
5. Place the glass slab back to its original position.
6. Fix a pin at O in contact with glass surface PQ.
7. Looking through the other side of the slab, fix two more pins such that these two pins and the image of the previous pin seem to be in a straight line.
8. Remove the pins and the glass slab and complete the ray diagram to get the apparant position of the image at I.
9. Measure OA and IA
10. Take three more observations with different vpositions of the pins each time.
PRECAUTIONS:
1. The glass slab must be correctly placed on the out line for all the observations.
2. Pins must be vertical.
3. While viewing through the other end of the glass slab, the feet of the pins must be adjusted to form a straight line and not the heads.
RESULT:
The refractive index of glass with respect to air is found to be _____
