Respuesta :
Explanation:
Formula which holds true for a leans with radii [tex]R_{1}[/tex] and [tex]R_{2}[/tex] and index refraction n is given as follows.
[tex]\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}][/tex]
Since, the lens is immersed in liquid with index of refraction [tex]n_{1}[/tex]. Therefore, focal length obeys the following.
[tex]\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}][/tex]
[tex]\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}][/tex]
and, [tex]\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}[/tex]
or, [tex]f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}[/tex]
[tex]f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}[/tex]
= 32.4 cm
Using thin lens equation, we will find the focal length as follows.
[tex]\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}[/tex]
Hence, image distance can be calculated as follows.
[tex]\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}[/tex]
[tex]s_{i} = \frac{fs_{o}}{s_{o} - f}[/tex]
[tex]s_{i} = \frac{32.4 \times 100}{100 - 32.4}[/tex]
= 47.9 cm
Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.
