Question

3. Convex lens of focal length 15 cm is made of material of refractive index 1.2. When placed in water (n=1.3), it will behave as [ ] a) converging lens of focal length 15 cm b) converging lens of focal length different them c) diverging lens of focal length 15 cm d) diverging lens of focal length different than 15 cm​

Answer 1

Given: f = 15 cm, µ(material) = 1.2 & µ(water) = 1.3.

Answer: (c) The lens will behave like a diverging lens of focal length 15 cm. (Actually it is - 15 cm as per sign convention.)

Explanation:

Keeping in mind that, light ray from rarer to denser medium bends towards the normal & from denser to rarer, it bends away from normal, we can proceed:-

Assume that the rays are parallel to te principal axis,

• Lens in air: __(see attachment 1)
• Lens in water: _(see attachment 2)

More:

Relationship between the focal length/power of a lens and thickness of it:-

• It a clear the more the refracting area, more refraction takes place.
• Therefore, if the the thickness of the refracting medium is increased, then there is more refraction. This implies that more the refraction, the less is its focal length.

$\rm Thickness \propto \dfrac{1}{Focal \ length}$

• And we can also write,

$\rm Thickness \propto Power.$

P.S.: *In attachment 1, the exemplar should be water drop in air.

Answer 2

Given: Convex lens of focal length 15 cm is made of material of refractive index 1.2. It is placed in water with refractive index= 1.3.

To find: Nature of lens

Solution: Let the focal length of the lens be f, refractive index of lens be ul, refractive index of medium be um and r1 and r2 be the radius of curvature of the lens.

The focal length is given by:

$\frac{1}{f} = ( \frac{ul}{um} - 1)( \frac{1}{r1} - \frac{1}{r2} )$

For air, um= 1

Therefore,

$\frac{1}{15} = (\frac{1.2}{1} - 1)( \frac{1}{r1} - \frac{1}{r2} )$

$\frac{1}{15} = 0.2( \frac{1}{r1} - \frac{1}{r2} )$

$\frac{1}{r1} - \frac{1}{r2} = \frac{1}{15 \times 0.2}$

$\frac{1}{r1} - \frac{1}{r2} = \frac{1}{3}$

Now, when kept in water, um= 1.3

The formula for focal length in this case:

$\frac{1}{f} = ( \frac{1.2}{1.3} - 1)( \frac{1}{r1} - \frac{1}{r2} )$

$\frac{1}{f} = ( \frac{1.2 - 1.3}{1.3} )( \frac{1}{3} )$

$\frac{1}{f} = ( \frac{ - 0.1}{1.3} )( \frac{1}{3} )$

$\frac{1}{f} = \frac{ - 1}{39}$

f = -39 cm

Negative focal length shows that lens behaves as a diverging lens.

Therefore, the lens will behave as option (d) diverging lens of focal length different than 15 cm.