To solve the problem, we need to find the focal length of a lens when the lens power is given.
1. Understanding the Relationship Between Power and Focal Length:
The power of a lens ($P$) is related to its focal length ($f$) by the formula:
$ P = \frac{1}{f} $
where $P$ is the power in diopters (D) and $f$ is the focal length in meters.
2. Given Information:
The power of the lens is given as 5D, so:
$ P = 5 \, \text{D} $
3. Finding the Focal Length:
Using the formula $ P = \frac{1}{f} $, we can rearrange it to find $f$:
$ f = \frac{1}{P} = \frac{1}{5} $ meters
$ f = 0.2 \, \text{m} = 20 \, \text{cm}$
Final Answer:
The focal length of the lens is $ {20 \, \text{cm}} $
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm. 