A glass sphere has radius R = 6.4 cm and index ofrefraction 1.7. A paperweight i

Question

A glass sphere has radius R = 6.4 cm and index ofrefraction 1.7. A paperweight is constructed by slicing through thesphere along a plane that is 2.0 cm from the center of the sphere,leaving height h = 4.4 cm. The paperweight is placed on atable and viewed from directly above by an observer who is distanced = 11 cm from the tabletop (figure below). When viewedthrough the paperweight, how far away does the tabletop appear tobe to the observer?

A glass sphere has radius R = 6.4 cm and index of refraction 1.7. A paperweight is constructed by slicing through the sphere along a plane that is 2.0 cm from the center of the sphere, leaving height h = 4.4 cm. The paperweight is placed on a table and viewed from directly above by an observer who is distanced = 11 cm from the tabletop (figure below). When viewed through the paperweight, how far away does the tabletop appear to be to the observer?

Explanation / Answer

We know that for refracting surfaces   n1/p     +   n2/i    = (n2 - n1)/r n1 = refractive index of the glass sphere = 1.7 n2 = refractive index of air =1 Here radius of curvature r = -6.4 cm (byapplying sign convertion) Here object distance p = 4.4cm Then 1.7 / 4.4 cm   +  1/i    = (1 - 1.6)/-6.4cm       1.6 /4.4cm     +   1/i   = 0.09 cm -1     0.39 cm -1   +   1/i   = 0.09 cm -1       i = -3.3 cm Now from the figure by noting the location of theobserver the table top is 9.7 cm away FROM THEOBSERVER.. Here observer's position is d=  11 cm Here i = -3.3 cm Thus table top is 3.3 cm below the surface of the sphere.THUS TABLE TOP IS  9.7 cm away FROM THE OBSERVER.       i = -3.3 cm Now from the figure by noting the location of theobserver the table top is 9.7 cm away FROM THEOBSERVER.. Here observer's position is d=  11 cm Here i = -3.3 cm Thus table top is 3.3 cm below the surface of the sphere.THUS TABLE TOP IS  9.7 cm away FROM THE OBSERVER.

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