A small pin fixed on a table top is viewed from above from a distance of 50 cm. By What distance would the pin appear to be raised , if it be viewed from the same point through a 15 cm thick glass slab held parallel to the table ? μ of glass is 1.5 . Does the answer depended on location of the slab ?
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Actual depth of the pin (d)=15 cm
Apparent depth of the pin (d1)=15 cm
Refractive index of the glass (μ) = 1.5
The ratio of actual depth to the apparent depth is equal to the refractive index of glass
μ =d/d1
or, d1 =d/ μ = 15/1.5 = 10 cm
The distance at with the pin appears to be raised = d1 - d = 15 - 10 = 5 cm
This small distance does not depend on the location of the slab.
Apparent depth of the pin (d1)=15 cm
Refractive index of the glass (μ) = 1.5
The ratio of actual depth to the apparent depth is equal to the refractive index of glass
μ =d/d1
or, d1 =d/ μ = 15/1.5 = 10 cm
The distance at with the pin appears to be raised = d1 - d = 15 - 10 = 5 cm
This small distance does not depend on the location of the slab.
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