A cube of edge 6 cm is placed over a printed page. at what distance from the top surface will the letters appear when seen from above? Refractive index of glass 1.5

Open in App

Solution

$d_{a p p e r e n t} = \frac{d_{a c t u a l}}{μ g_{1} a} = \frac{6 c m}{1.5} = 4 c m$
So letters will appear at 4 cm from top.

Suggest Corrections

Similar questions

A printed page of is kept pressed by a glass cube (n=1.5) of edge 9 cm. By what amount will the printed letters appear to be shifted when viewed from the top?

Q. A printed page is kept pressed by a glass cube (

μ

= 1.5) of edge 6.0 cm. By what amount will the printed letters appear to be shifted when viewed from the top?

Q. A printed page is kept pressed by a glass cube (

μ

= 1.5) of edge 6.0 cm. By what amount will the printed letters appear to be shifted when viewed from the top?

A printed page is kept pressed by a glass cube ( $μ$ = 1.5) of edge 6.0 cm. By what amount will the printed letters appear to be shifted when viewed from the top?

There is a black spot on a table. A glass slab of thickness 6 cm is placed on the table over the spot. The refractive index of glass is $\frac{3}{2}$ . At what depth from the upper surface will the spot appear when viewed from above?

Join BYJU'S Learning Program

A cube of edge 6 cm is placed over a printed page. at what distance from the top surface will the letters appear when seen from above? Refractive index of glass 1.5

dapperent=dactualμg1a=6cm1.5=4cm So letters will appear at 4 cm from top.

$d_{a p p e r e n t} = \frac{d_{a c t u a l}}{μ g_{1} a} = \frac{6 c m}{1.5} = 4 c m$
So letters will appear at 4 cm from top.