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.
![\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7Bs_%7Bo%7D%7D%20%2B%20%5Cfrac%7B1%7D%7Bs_%7Bi%7D%7D)
Hence, image distance can be calculated as follows.
![s_{i} = \frac{fs_{o}}{s_{o} - f}](https://tex.z-dn.net/?f=s_%7Bi%7D%20%3D%20%5Cfrac%7Bfs_%7Bo%7D%7D%7Bs_%7Bo%7D%20-%20f%7D)
![s_{i} = \frac{32.4 \times 100}{100 - 32.4}](https://tex.z-dn.net/?f=s_%7Bi%7D%20%3D%20%5Cfrac%7B32.4%20%5Ctimes%20100%7D%7B100%20-%2032.4%7D)
= 47.9 cm
Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.
Answer:
![t=750s](https://tex.z-dn.net/?f=t%3D750s)
Explanation:
The two cars are under an uniform linear motion. So, the distance traveled by them is given by:
![\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt](https://tex.z-dn.net/?f=%5CDelta%20x%3Dvt%5C%5Cx_f-x_0%3Dvt%5C%5Cx_f%3Dx_0%2Bvt)
is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:
![x_0_1=6km\\x_0_2=0](https://tex.z-dn.net/?f=x_0_1%3D6km%5C%5Cx_0_2%3D0)
We have
. Thus, solving for t:
![x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s](https://tex.z-dn.net/?f=x_0_1%2Bv_1t%3Dx_0_2%2Bv_2t%5C%5Cx_0_1%3Dt%28v_2-v_1%29%5C%5Ct%3D%5Cfrac%7Bx_0_1%7D%7Bv_2-v_1%7D%5C%5Ct%3D%5Cfrac%7B6%2A10%5E3m%7D%7B28%5Cfrac%7Bm%7D%7Bs%7D-20%5Cfrac%7Bm%7D%7Bs%7D%7D%5C%5Ct%3D750s)