Physics

We know, $vμ_{1} −uμ_{2} =Rμ_{2}−μ_{1} $

For 1, we get:

$vμ_{2} −−dμ_{1} =R(μ_{2}−μ_{1}) $ , $−d=u=$ object distance

or, $vμ_{1} =Rμ_{2}−μ_{1} −dμ_{1} =t$

For 2, we get:

$dμ_{1} −(2R−v)μ_{2} =Rμ_{2}−μ_{1} $

or, $(Rμ_{2}−μ_{1} −dμ_{1} )=(2R−v)μ_{2} =vμ_{2} $

or, $2R−v=v$

or, $R=v$

So, $dμ_{1} =rμ_{2}−μ_{1} $

or, $d=(μ_{2}−μ_{1}μ_{1} )R$

Let,

$n_{1}$ - refractive index of medium from which rays incident.

$n_{2}$ - refractive index of another medium.

u - distance of object from pole of spherical surface

v - distance of image from pole of spherical surface

$tanα=OMMN $

$tanγ=MCMN $

$tanβ=MIMN $

Now, for $Δ$NOC, i is the exterior angle.

$i=∠NOM+∠NCM$

$i=OMMN +MCMN $.........(1)

Similarly,

$r=MCMN −MIMN $.........(2)

Now by using snells law we get

$n_{1}sini=n_{2}sinr$

Or for small angles

$n_{1}i=n_{2}r$

Substituting i and r from eq. (1) and (2), we get

$OMn_{1} +MIn_{2} =MCn_{2}−n_{1} $

As,

OM = -u, MI = +v, MC = + R

Hence equation becomes

$vn_{2} −un_{1} =Rn_{2}−n_{1} $

Solution:

Let image distance be $v$, object distance be $−u$ and focal length be $f$.

Let distance between object and image be $s$.

$∴s=u+v$

Now, $v1 +u1 =f1 $

or, $uvu+v =f1 $

or, $u(s−u)s =f1 $

or, $s=u−fu_{2} $

For point of minima, $duds =0$

or, $(u−f)_{2}u_{2}−2uf =0$

or, $u=0,2f$

Therefore, $s=2f−f(2f)_{2} =4f$

$m=OI =uv $

Note:

Sign convention must be followed while using formula for magnification. Hence, it can be positive or negative.

$∣m∣>1⟹$ image is magnified.

$∣m∣=1⟹$ image is same size as object.

$∣m∣<1⟹$ image is diminished.

Solution:

When the curved surface of the lens (refractive index $μ$) is in contact with the table, the image of the bottom-most point of lens (in glass) is formed due to refraction at plane face.

The image of O appears at $I_{1}$.

Here, $u_{1}=AO=−4cm,v_{1}=AI_{1}=3cm$, $μ_{1}=μ$, $μ_{2}=1$, and $R_{1}=∞$

$∴v_{1}μ_{2} −u_{1}μ_{1} =R_{1}μ_{2}−μ_{1} $ gives,

$−31 −−4μ =∞1−μ $................ (i)

When the plane surface of the lens in contact with the table, the image of center of the plane face is formed due to refraction at curved surface. The image of O is formed at $I_{2}$.

Here, $u_{2}=AO=−4cm,v_{2}=AI_{2}=−825 cm$, $μ_{1}=μ,μ_{2}=1$, and $R_{2}=−R$

$∴v_{2}μ_{2} −u_{2}μ_{1} =R_{2}μ_{2}−μ_{1} $ gives.

$(−825 )1 −−4μ =−R1−μ $

From Eq. (i), we get:

$μ=4/3$, therefore this equation gives

$−258 +44/3 =−R(1−34 ) =−258 +31 =3R1 $

or $751 =3R1 $

This gives $R=25cm$

The focal length ($f$) of plano-convex lens $(R_{1}=R$ and $R_{2}=∞)$ is $f1 =(μ−1)(R1 −∞1 )=Rμ−1 =2534 −1 =751 $

$⇒f=75cm$

The focal length of a thin biconvex lens is $20cm$. When an object is moved from a distance of $25cm$ in front of it to $50cm$, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $m_{50}m_{25} $ is :

When object is at $25cm$

$u=−25cm$, $v=100cm$

$m_{25}=uv =25100 $.....(1)

When object is at $50cm$

$u=−50cm$, $v=100/3cm$

$m_{50}=uv =150100 $.....(2)

We have to find $m_{50}m_{25} $

$m_{50}m_{25} =150100 25100 =6$

Distance between object and image $=2f+2f=4f$