Explanation:
Formula which holds true for a leans with radii 
and 
and index refraction n is given as follows.
![\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%28n%20-%201%29%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
Since, the lens is immersed in liquid with index of refraction 
. Therefore, focal length obeys the following.
![\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7B1%7D%7D%20%3D%20%5Cfrac%7Bn%20-%20n_%7B1%7D%7D%7Bn_%7B1%7D%7D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
![\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%28n%20-%201%29%7D%20%3D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
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.
Answer:
true
Explanation:
this the nucleus is located at the centre and contains protons and neutrons
Answer:
L = 1.15 m
Explanation:
The diffraction phenomenon is described by the equation
a sin θ = m λ
Where a is the width of the slit, λ the wavelength and m is an integer, the order of diffraction is left.
The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry
tan θ = y / L
tan θ = sint θ / cos θ≈ sin θ
We substitute in the first equation
a (y / L) = m λ
The first maximum occurs for m = 1
The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is
y = 1.15 / 2 = 0.575 cm
y = 0.575 10⁻² m
Let's clear the distance to the screen (L)
L = a y / λ
Let's calculate
L = 115 10⁻⁶ 0.575 10⁻² / 575 10⁻⁹
L = 1.15 m
gravitational force of planet exerted on its object near the surface is known as weight
so here we know that gravitational force of mars is much less than the gravitational force of Earth
So on the surface of mars the Weight of objects must be much less than the weight of object on surface of Earth
so here correct answer must be
<em>D. Your weight would decrease.</em>
