A convex lens of refractive index 1.5 and focal length $f = 18$ cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $α \times f$ . The value of $α$ is ____.

(Refractive index of water = $\frac{4}{3}$ ) [2026]

Question

A convex lens of refractive index 1.5 and focal length $f = 18$ cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $α \times f$ . The value of $α$ is ____.

(Refractive index of water = $\frac{4}{3}$ ) [2026]

Smart Achievers · Accepted Answer

(3)

$\frac{1}{f_{air}} = (\frac{1.5 - 1}{1}) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\frac{1}{f_{water}} = (\frac{1.5 - \frac{4}{3}}{\frac{4}{3}}) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\Rightarrow \frac{f_{water}}{f_{air}} = \frac{0.5}{0.5 / 4} = 4$

$\Rightarrow f_{water} - f_{air} = 3 f$

Solution

Q.
A convex lens of refractive index 1.5 and focal length $f = 18$ cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $α \times f$ . The value of $α$ is ____.

(Refractive index of water = $\frac{4}{3}$ ) [2026]

Ans.
(3)

$\frac{1}{f_{air}} = (\frac{1.5 - 1}{1}) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\frac{1}{f_{water}} = (\frac{1.5 - \frac{4}{3}}{\frac{4}{3}}) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\Rightarrow \frac{f_{water}}{f_{air}} = \frac{0.5}{0.5 / 4} = 4$

$\Rightarrow f_{water} - f_{air} = 3 f$

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Solution

Q. A convex lens of refractive index 1.5 and focal length f=18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α×f. The value of α is ____. (Refractive index of water = 43) [2026]

Ans. (3) 1fair=(1.5-11)(1R1+1R2) 1fwater=(1.5-4343)(1R1+1R2) ⇒fwaterfair=0.50.5/4=4 ⇒fwater-fair=3f

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Q.
A convex lens of refractive index 1.5 and focal length $f = 18$ cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $α \times f$ . The value of $α$ is ____.

(Refractive index of water = $\frac{4}{3}$ ) [2026]

Ans.
(3)

$\frac{1}{f_{air}} = (\frac{1.5 - 1}{1}) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\frac{1}{f_{water}} = (\frac{1.5 - \frac{4}{3}}{\frac{4}{3}}) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\Rightarrow \frac{f_{water}}{f_{air}} = \frac{0.5}{0.5 / 4} = 4$

$\Rightarrow f_{water} - f_{air} = 3 f$