Answer:
T = 4.905[N]
Explanation:
In order to solve this problem we must perform a sum of forces on the vertical axis.
∑Fy = 0
We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.
![T-m*g=0\\T=0.5*9.81\\T=4.905[N]](https://tex.z-dn.net/?f=T-m%2Ag%3D0%5C%5CT%3D0.5%2A9.81%5C%5CT%3D4.905%5BN%5D)
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Explanation:
Formula which holds true for a leans with radii
and
and index refraction n is given as follows.
Since, the lens is immersed in liquid with index of refraction
. Therefore, focal length obeys the following.
and,
or,
= 32.4 cm
Using thin lens equation, we will find the focal length as follows.

Hence, image distance can be calculated as follows.


= 47.9 cm
Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.