Answer:
Explanation:
a )
from lens makers formula

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face
putting the values

1.462 = 2 - 1 / r₂
1 / r₂ = .538
r₂ = 1.86 cm .
= 18.6 mm .
b )
object distance u = 25 cm
focal length of convex lens f = 1.8 cm
image distance v = ?
lens formula



.5555 - .04
= .515
v = 1.94 cm
c )
magnification = v / u
= 1.94 / 25
= .0776
size of image = .0776 x size of object
= .0776 x 10 mm
= .776 mm
It will be a real image and it will be inverted.
The presence of helium gas indicates the radioactive sample is most likely decaying by α-decay, or alpha decay. α-decay is the type of radioactive decay in which an atomic nucleus emits α particles. α particles are Helium nuclei. So the correct answer would be α-decay.
Answer:
Light is one of nature's ways of moving energy from one place to another.
Explanation:
It has no substance, or mass. How does light travel? Light behaves like a traveling wave, something like waves in a string or on the surface of water.
Answer:
Explanation:
Given
Initial Intensity of light is S
when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.
When it is passed through a second Polarizer with its transmission axis 

here 


When it is passed through third Polarizer with its axis
to first but
to second thus 



When middle sheet is absent then Final Intensity will be zero
Answer:
2.6 m
Explanation:
The work done by the bird is given by

where
F is the force exerted
d is the distance covered
In this problem, we know:
is the work
is the force
Solving the equation for d, we find the distance covered by the bird:
