A piece of written material appears raised by 7 mm when placed under a rectangular glass block of refractive index 1. 5. Find the thickness of the glass block.

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Solution

Refractive index =1.5
Let real depth of glass block be 'x' mm

The apparent depth will be (x-7) mm

We know,
refractive index,
m=Real depth/Apparent depth

1.5=x/x-7
1.5(x-7)=x
1.5x-10.5=x
0.5x=10.5
x=21 mm

Hence thickness of glass block =21 mm

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A piece of written material appears raised by 7 mm when placed under a rectangular glass block of refractive index 1. 5. Find the thickness of the glass block.

Refractive index =1.5 Let real depth of glass block be 'x' mm The apparent depth will be (x-7) mm We know, refractive index, m=Real depth/Apparent depth 1.5=x/x-7 1.5(x-7)=x 1.5x-10.5=x 0.5x=10.5 x=21 mm Hence thickness of glass block =21 mm

Refractive index =1.5
Let real depth of glass block be 'x' mm

The apparent depth will be (x-7) mm

We know,
refractive index,
m=Real depth/Apparent depth

1.5=x/x-7
1.5(x-7)=x
1.5x-10.5=x
0.5x=10.5
x=21 mm

Hence thickness of glass block =21 mm