An air bubble in glass refracting 1.5 is situated at a distance 3 cm from a convex surface of diameter 10 cm as shown the distance from surface at which the image of bubble appears is
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Answered by
42
Hi there.
Refractive Index of the glass(μ₁) = 1.5
Refractive Index of the air(μ₂) = 1
Diameter of the Convex Surface(d) = 10 cm.
∴ Radius of the Convex surface(r) = 5 cm. (negative)
Real Depth of the Air Bubble (u) = 3 cm. (negative)
Let the Apparent Depth of the Air bubble (or the distance of the image of the bubble from the air surface) = v cm.
∵ When the light from the bubble goes from the glass to the air, then
μ₂/v - μ₁/u = (μ₂ - μ₁)/r
∴ 1/v - 1.5/-3 = (1.5 - 1)/-5
⇒ 1/v + 1/2 = 0.5/-5
⇒ 1/v = -1/2 - 1/10
⇒ 1/v = (-5 - 1)/10
⇒ 1/v = -6/10
∴ v = -10/6 cm.
∴ v = -5/3
∴ v = -1.67 cm
Hence, the image is formed at the distance of 1.67 cm from the convex surface.
Hope it helps.
Refractive Index of the glass(μ₁) = 1.5
Refractive Index of the air(μ₂) = 1
Diameter of the Convex Surface(d) = 10 cm.
∴ Radius of the Convex surface(r) = 5 cm. (negative)
Real Depth of the Air Bubble (u) = 3 cm. (negative)
Let the Apparent Depth of the Air bubble (or the distance of the image of the bubble from the air surface) = v cm.
∵ When the light from the bubble goes from the glass to the air, then
μ₂/v - μ₁/u = (μ₂ - μ₁)/r
∴ 1/v - 1.5/-3 = (1.5 - 1)/-5
⇒ 1/v + 1/2 = 0.5/-5
⇒ 1/v = -1/2 - 1/10
⇒ 1/v = (-5 - 1)/10
⇒ 1/v = -6/10
∴ v = -10/6 cm.
∴ v = -5/3
∴ v = -1.67 cm
Hence, the image is formed at the distance of 1.67 cm from the convex surface.
Hope it helps.
Answered by
24
Answer:
the answer is minus 2.5
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