A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to

<p>Using lens formula,</p> <p>For Ist case : $${1 \over f} = \left( {{a^{{\mu _g}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p> <p>where, ${R_1},{R_2}$ = radii of refracting surfaces</p> <p>$f$ = focal length</p> <p>${a^{{\mu _g}}} = {{^{{\mu _g}}} \over {^{{\mu _a}}}} = {{1.5} \over 1} = 1.5$</p> <p>So, ${1 \over {24}} = (1.5 - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ (As f = 24 cm given)</p> <p>$$ \Rightarrow {1 \over {24}} = 0.5\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p> <p>$$ \Rightarrow {1 \over {12}} = \left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p> <p>$\Rightarrow {1 \over {{R_1}}} - {1 \over {{R_2}}} = {1 \over {12}}$ .... (i)</p> <p>For IInd case:</p> <p>$${1 \over f} = \left( {w{\mu _g} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p> <p>$$ \Rightarrow {1 \over f} = \left( {{{1.5} \over {1.33}} - 1} \right)\left( {{1 \over {12}}} \right)$$ ..... from (i)</p> <p>$$ \Rightarrow {1 \over f} = {{17} \over {133}} \times {1 \over {12}} \Rightarrow f = {{133 \times 12} \over {17}}$$</p> <p>$\Rightarrow f = 93.88$ cm</p> <p>$\approx$ 94 cm</p> <p>Hence, option 1 is correct.</p>