Question

5. A double convex lens of focal length 6cm is
made of glass of refractive index 1.5. the radius of curvature of one surface is double that of the other surface .the value of small radius of curvature
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Answer 1

GiVen:

• Focal length = 6cm
• Refractive index = 1.5

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AnSwer:

As, radius of curvature of one surface is double that of the other surface,

• R1 = R and R2 = -2R

Now,

From the lens formula,

$\sf \implies \: \dfrac{1}{f} = (μ - 1)( \dfrac{1}{R1 } - \dfrac{1}{R2} )$

$\sf \implies \: \dfrac{1}{6} = (1.5 - 1)( \dfrac{1}{R} - \dfrac{1}{2R}) \\ \\ \sf \implies \: \frac{1}{6} = \frac{0.5 \times 3}{2R} \\ \\ \sf \implies \: R= 6(0.5 \times 3) \\ \\ \sf \implies \:R = 4.5 \: cm$

Therefore, value of small radius is 4.5cm.

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Answer 2

please mark this answer as brainliest.