Question 107: Ray Optics

Question

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

Accepted Answer

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

Answer

96 cm

Answer

72 cm

Answer

24 cm

Answer

48 cm

Ray Optics

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

Choose the correct answer: