The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

Here, \(\lambda =5000\quad \overset { \circ }{ A } =5\times 10^{ -7 }m\)
\(v=\frac { c }{ \lambda } =\frac { 3\times 10^{ 8 } }{ 5\times 10^{ -7 } } =6\times 10^{ 14 }\) hertz
On reflection, there is no change in wavelength or frequency. Therefore,
\(\lambda '=\lambda =5000\quad \overset { \circ }{ A } \quad ;\quad v'=v=6\times 10^{ 14 }Hz.\)
For reflected ray to be normal to incident ray,
\(i+r=90°\) or \(i+i=90°\)   \((\because r=i)\)
\(\therefore \) \(i=90/2=45°\)​​​

Differences between interference and diffraction

A = 60⁰ ,  \(\delta \)_m = 30⁰
i = e = ¾ A = 450
as A + \(\delta \) = i + e
60 + \(\delta \) = 45 +45
or \(\delta \) = 30⁰
Refractive index,
\(\mu \) = sin a + \(\delta \)m /2/sin A/2 = sin 60⁰+30⁰/2/sin 60⁰/2
= sin 45⁰/sin30⁰ = 1\(\surd 2\) 1/2 = \(\surd 2\) = 1.414

A hollow prism contains air which does not cause dispersion. The faes AB and AC of the hollow prism behave like parallel sides of glass plates. The beam is laterally deviated at each of the two refracting faces. However, the rays of different colours emerge parallel to each other. So there is no dispersion.

Real depth = y cm
Apparent depth = y- x cm
Refractive index of oil,
\({ \mu }\) = real depth/ apparent depth = y / y-x

Shrinks

Violet

\({ \mu }_{ 0 }={ \mu }_{ e }\)

Hypermetropia is long sightedness.It may be due to contraction in size of the eye ball or increase in focal length of eye lens, when fully relaxed.Presbyopia is long sightedness due to old age when eye ball has normal length.Only the focal length of eye lens in fully relaxed position has increased.

Increase in focal length of eye lens when eye is fully relaxed.

These rays are focussed in front of the retina.

d = 25c.Yes

Range of the vision is the distance between near point and far point of eye.

It is the ability of the eye to observe distinctly the objects situated at widely different distance from the eye.

It is a spot at about the centre of the retina, which is most sensitive to light.