An air bubble in a glass slab $μ = 1.5$ is 5 cm deep when viewed from one face and 2cm deep when viewed from the opposite face. The thickness of the glass slab is

10.5 cm

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7 cm

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10 cm

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7.5 cm

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Solution

The correct option is A
10.5 cm

Real height = $h r$

Apparent height = $h a$

$μ = \frac{h r_{1}}{h a_{1}}; μ = \frac{h r_{2}}{h a_{2}} t = h r_{1} + h r_{2} = μ [h a_{1} + h a_{2}] = 1.5 [5 + 2] = 1.5 \times 7 t = 10.5 c m$

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An air bubble in a glass slab μ=1.5 is 5 cm deep when viewed from one face and 2cm deep when viewed from the opposite face. The thickness of the glass slab is

The correct option is A 10.5 cm Real height = hr Apparent height = ha μ=hr1ha1;μ=hr2ha2t=hr1+hr2=μ[ha1+ha2]=1.5[5+2]=1.5×7t=10.5cm

An air bubble in a glass slab $μ = 1.5$ is 5 cm deep when viewed from one face and 2cm deep when viewed from the opposite face. The thickness of the glass slab is

The correct option is A
10.5 cm

Real height = $h r$

Apparent height = $h a$

$μ = \frac{h r_{1}}{h a_{1}}; μ = \frac{h r_{2}}{h a_{2}} t = h r_{1} + h r_{2} = μ [h a_{1} + h a_{2}] = 1.5 [5 + 2] = 1.5 \times 7 t = 10.5 c m$