A double convex lens made of a material of refractive index 1-5 and having a focal length of 10 cm is immersed in liquid of refractive index 3-0 - The lens will behave asA- Diverging lens of focal length 10 cmB- Diverging lens of focal length 10-3 cmC- Converging lens of focal length 30 cmD- Converging lens of focal length 10-3 cm

The correct option is A Diverging lens of focal length 10 cmUse this formula, f_i/f_a=_aμ_g 1/_lμ_g 1⇒f_l/+10=1.5 1/1.5/3 1⇒ f_l = 10 cm (i.e., diverging lens) ...

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Question

A double convex lens made of a material of refractive index 1.5 and having a focal length of 10 cm is immersed in liquid of refractive index 3.0. The lens will behave as

Diverging lens of focal length 10 cm

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Diverging lens of focal length

\frac{10}{3} c m

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Converging lens of focal length

\frac{10}{3} c m

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Converging lens of focal length 30 cm

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Solution

The correct option is A Diverging lens of focal length 10 cm
Use this formula,
$\frac{f_{i}}{f_{a}} = \frac{_{a} μ_{g} - 1}{_{l} μ_{g} - 1} \Rightarrow \frac{f_{l}}{+ 10} = \frac{1.5 - 1}{\frac{1.5}{3} - 1} \Rightarrow f_{l} = - 10 c m$ (i.e., diverging lens)

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A double convex lens made of a material of refractive index 1.5 and having a focal length of 10 cm is immersed in liquid of refractive index 3.0. The lens will behave as

The correct option is A Diverging lens of focal length 10 cmUse this formula, fifa=aμg−1lμg−1⇒fl+10=1.5−11.53−1⇒fl=−10cm (i.e., diverging lens)

The correct option is A Diverging lens of focal length 10 cm
Use this formula,
$\frac{f_{i}}{f_{a}} = \frac{_{a} μ_{g} - 1}{_{l} μ_{g} - 1} \Rightarrow \frac{f_{l}}{+ 10} = \frac{1.5 - 1}{\frac{1.5}{3} - 1} \Rightarrow f_{l} = - 10 c m$ (i.e., diverging lens)