For v1,
\([\frac{1}{v_{1}}+\frac{1}{-10}=\frac{1}{12}=\frac{1}{v}=-\frac{1}{12}+\frac{1}{10}=\frac{-10+12}{120}=\frac{2}{120}=\frac{1}{60}]\)
v1=60cm
For v2,
\(\frac{1}{v_{1}}+\frac{1}{-80}=\frac{1}{10}=\frac{1}{v_{2}}=\frac{1}{10}-\frac{1}{80}=\frac{8-1}{80}=\frac{7}{80}\)
\(v_{2}=\frac{80}{7}cm\)
Reflection from M1
\(\frac{1}{v_{1}}+\frac{1}{-10}=\frac{1}{f_{1}}=\frac{1}{12}\)
\(\frac{1}{v}+\frac{1}{-10}=\frac{1}{f}=\frac{1}{12}=\frac{1}{v_{1}}+\frac{1}{-80}\)
\(\frac{1}{v_{2}}=\frac{1}{10}=\frac{1}{v}-\frac{1}{80}\)
For lens
\(v=\frac{uf}{u+f}=\frac{(-80)10}{-80+10}=\frac{80}{7}cm\)
For an image to form at S itself Ray after reflection from mirror 'M2 must be PARALLEL to the lens
L must form an image at the focus of 'M2'.
\(\frac{n}{7}=10+\frac{80}{7}=\frac{150}{7}\)
n=150
If l2 is formed at the pole of M1,
Then, \(\frac{n}{7}=\frac{80}{7}\)
n = 80
And if l2 is formed at the center of curvature of M1,
\(\frac{n}{7}=\frac{80}{7}+20\)
\(\therefore n=220\)
So, the answer is 80 or 150 or 220
A current element X is connected across an AC source of emf \(V = V_0\ sin\ 2ΟΞ½t\). It is found that the voltage leads the current in phase by \(\frac{Ο}{ 2}\) radian. If element X was replaced by element Y, the voltage lags behind the current in phase by \(\frac{Ο}{ 2}\) radian.
(I) Identify elements X and Y by drawing phasor diagrams.
(II) Obtain the condition of resonance when both elements X and Y are connected in series to the source and obtain expression for resonant frequency. What is the impedance value in this case?
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Optics, deals with the determination of behaviour and the properties of light, along with its interactions with the matter and also with the instruments that are used to detect it.
Ray optics is also known as the geometrical optics and it is a branch of science which describes light propagation.
Reflection is the change in direction of light at an interface in-between two different media so that the wave-front returns into a medium from which it was originated.
Speed of light is the rate at which the light travels in free space.
A phenomenal change in image formed when the light is passed from one medium to another which is called Refraction.
Total Internal Reflection is the reflection of light when the light ray enters into a rarer medium from a denser medium and the angle of incidence is higher than the critical angle of incidence then that light ray will be reflected back to the denser medium.
Read More: Ray Optics and Optical Instruments