\(\frac{1}{f}  =(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)  \)
\(=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{(-10)}\right)\)
(Since plano convex lens)
\(=0.5\left[\frac{1}{10}\right]=\frac{1}{20}  \)
\(\mathrm{f}_t  =20 \mathrm{~cm}\)
Formula for silvered lenses
\(\frac{1}{\mathrm{~F}}  =\frac{2}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_m}  \)
\(\frac{1}{\mathrm{~F}} =\frac{2}{20}+\frac{1}{\infty}  \)
\(\therefore \mathrm{F}  =\frac{20}{2}=10 \mathrm{~cm}\)

\(\frac{I}{f}=\left(\frac{\mu_{\mathrm{L}}}{\mu_L}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
When the biconvex lens of glass dipped in liquid, it acts as a plane sheet of glass.
\(\therefore \mathrm{f}=\infty, \frac{1}{\mathrm{f}}=0 \quad \frac{\mu_g}{\mu_{\mathrm{L}}}-1=0 ; \frac{\mu_{\mathrm{s}}}{\mu_{\mathrm{L}}}=1, \mu_{\mathrm{s}}=\mu_{\mathrm{L}}\)

At end A,
\(\frac{1}{f}  =\frac{1}{u_A}+\frac{1}{v_A}  \)
\(\therefore \frac{1}{v_A}  =\frac{1}{-10}-\frac{1}{-20}  \)
\(\frac{1}{v_A}  =-\frac{1}{10}+\frac{1}{20}=\frac{-2+1}{20}=-\frac{1}{20}  \)
\(v_A  =-20 \mathrm{~cm} \)
\(\left|v_{\wedge}\right|=20 \mathrm{~cm}\)
At end B,
\(\frac{1}{f}  =\frac{1}{u_B}+\frac{1}{v_B}  \)
\(\frac{1}{v_B} =\frac{1}{f}-\frac{1}{u_B},  \)
\(u_B  =-30 \mathrm{~cm}  \)
\(\frac{1}{v_B}  =-\frac{1}{10}+\frac{1}{30}  \)
\(=\frac{-3+1}{30}=\frac{-2}{30}=\frac{-1}{15}  \)
\(v_B  =-15 \mathrm{~cm}  \)
\(\left|v_B\right|  =15 \mathrm{~cm} \)
\(\therefore \quad\left|\mathrm{v}_{\mathrm{A}}\right|-\left|\mathrm{v}_{\mathrm{B}}\right| \) is the length of the image
= 20 - 15 = 5 cm

v = nג
In an isotropic medium, there is no change in the frequency of the light. So, the speed of light depends on wavelength of light.

Real image :
This type of image which can be formed on a screen and can also be seen with the eyes is called real image.
Virtual image :
Image which cannot be formed on the screen but can only be seen with the eyes.

The bouncing back of light into the same medium when it encounters a reflecting surface is called reflection of light.

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(m=\frac{-v}{u}=-4, f=20 \mathrm{~cm} \text { (Given) } \)
V = 4u
\(\frac{1}{f} =\frac{1}{4 u}-\frac{1}{u} \)
\(=\frac{1-4}{4 u}=\frac{-3}{4 u} \)
\(\frac{1}{f} =\frac{-3}{4 u} \)
4u = -3 x f
\(u=\frac{-3}{4} \times 20=-15 \mathrm{~cm}\)

The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

The angle between the incident and deviated light is called Angle of deviation due to refraction. When light travels from
(i) rarer to denser medium, d = i - r
(ii) denser to rarer medium, d = r - i

The law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the mirror all lie in the same plane. The angle of reflection is equal to the angle of incidence.

(i) Consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY as shown in Figure.

(ii) The incident wavefront is AB and the reflected wavefront is A'B' in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L', M' respectively.
(iii) By the time point A of the incident wavefront touches the reflecting surface, point B is yet to travel a distance BB' to touch the reflecting surface a B'.
(iv) When point B falls on the reflecting surface at H', point A would have reached A.
(v) This is applicable to all the points on the wavefront. Thus, the reflected wavefront A'B' emanates as a plane wavefront. The two normals Nand N' are considered at the points where the rays Land Mfallon the reflecting surface.
(vi) As reflection happens in the same medium, the speed of light is the same before and after the reflection.
(vii) Hence, the time is taken for the ray to travel from B to B' is the same as the time taken for the ray to travel from A to A'.
(viii) Thus, the distance BB' is equal to the distance AA'; (A~A' = BB').
(a) The incident rays, the reflected rays, and the normal are in the same plane.
(b) Angle of incidence,\(\angle i=\angle NAL={ 90 }^{ o }-\angle NAB=\angle BAB'\) 
Angle of reflection,
∠r= ∠N' B' M' = 900 - ∠N' B' A' = A' B' A' 
(ix) For the two right-angle triangles, ΔABB' and ΔB' A' A', the right angles, ∠B and ∠A' are equal, (∠B and∠A = 900); the two sides, ∠A' and ∠B' are equal, (AA'= BB'); the side AB' is common.
(x) Thus, the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB' and A' B' A' must also be equal.
i = r
Hence, the laws of reflection are proved.

\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \cfrac { 2 }{ f } \right) \) 
\(\cfrac { 1 }{ 2 } =\left( \mu -1 \right) \) 
\(\mu =1.5\)

Mirror Equation :

(i) AB is an object which is placed on the principal axis of a concave mirror beyond the center of curvature C. A' B' is an image which is formed between the point pole P, and the centre of curvature.
(ii) From the figure As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B'PA'.
(iii) The triangles ∠BPA and ∠B'PA' are similar. Thus, from the rule of similar triangles,
\(\cfrac { { A }^{ ' }{ B }^{ ' } }{ AB } =\cfrac { { PA }^{ ' } }{ PA } \)             ................(1)
(iv) The other set of similar triangles are, ΔDPF and ΔB'A'F. (PD is almost a straight vertical line)
\(\cfrac { { A }^{ ' }B' }{ PD } =\cfrac { A'F }{ PF } \) 
(v) As, PD = AB the above equation becomes,
\(\cfrac { A'B' }{ AB } =\cfrac { A'F }{ PF } \)            ......(2)
(vi) From equations (1) and (2) we can write,
\(\cfrac { PA' }{ PA } =\cfrac { A'F }{ PF } \) 
(vii) As, A'F = PA' - PF, the above equation becomes,
\(\cfrac { PA' }{ PA } =\cfrac { PA'-PF }{ PF } \)            .....(3)
(viii) We can apply the sign conventions for the various distances in the above equation
PA = - u, PA' = -v, PF = - f
(ix) All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
\(\cfrac { -v }{ -u } =\cfrac { -v-\left( -f \right) }{ -f } \) 
On further simplification,
\(\cfrac { v }{ u } =\cfrac { v-f }{ f } ;\cfrac { v }{ u } =\cfrac { v }{ f } -1 \) 
Dividing either side with v,
 \(\cfrac { 1 }{ u } =\cfrac { 1 }{ f } -\cfrac { 1 }{ v } \)
After rearranging,
\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
The above equation is called mirror equation.
Lateral magnification:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
Magnification (m) \(=\frac{\text { height of the image }\left(h^{\prime}\right)}{\text { height of the image }(h)} \)
\(m=\frac{h^{\prime}}{h} \)  ....(1)
Applying proper sign conventions for equation,
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A} \)
\(A^{\prime} B^{\prime}=-h^{\prime}, A B=h, P A^{\prime}=-v, P A=-u \)
\(-\frac{h}{h}=\frac{-v}{-u} \)
On simplifying we get,
\(\mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{\mathrm{v}}{\mathrm{u}}\)  ...(2)
Using mirror equation, we can further write the magnification as,
\(m=\frac{h^{\prime}}{h}=\frac{f-v}{f}=\frac{f}{f-u}\)   ..(3)

Convergent lens - Virtual image

(c) Assertion is true but Reason is false

I and IV only

 Dispersive power is always negative