Question

A postage stamp kept below a rectangular glass slab of refractive index 1.5 when viewed from vertically above it appears to be raised by 7 .0 mm calculate the thickness of the glass slab

Answer 1

》t(1-1/u)=7

》t(1-1/1.5)=7

》t(1.5/1.5-1/1.5)=7

》t(.5÷1.5)=7

》t(1/3)=7

》t=7 × 3

》t=21mm

Answer 2

$\huge\mathrm{Answer}$

Given ⚡

Refractive index ( μ ) = 1.5

Shift = 7 mm

Solution ⚡

Let real depth be - x mm

Shift = Real depth × ( 1 - 1 / μ of glass slab )

➡ Shift = x × ( 1 - 1 / 1.5 )

➡ Shift = x × ( ( 1.5 - 1 ) / 1.5 )

➡ Shift = x × 0.5 / 1.5

➡ 7 = x × 5 / 15 [After removing the decimal points ]

➡ x = ( 7 × 15 ) / 5

➡ x = 7 × 3

➡ x = 21 mm = 2.1 cm

Therefore, real depth is 2.1 cm or 21 mm

Notes ⭐

➡ An object placed in a denser medium when viewed from a real medium appears to be at a depth lesser than its real depth this is because of the refraction of light

➡ Shift = Real depth - Apparently depth

➡ Refractive index of a medium = Real depth / Apparently depth