In the diagram shown, a monochromatic light is placed in front of polaroid (I) as shown nelow.

As, per the given question, monochromatic light emerging from polaroid I is plane polarised. When polaroid II is placed in front of this polaroid II, then I and II are set in crosses positions, i.e. pass axes of I and II are at \({ 90 }^{ \circ }\)

When a polaroid III is placed in between I and II, no light will emerge from II, if pass axis of III is parallel to that od I or II. In all the other case, light will emerge from II.

From the given figure ,of two slit interference arrangement,we can write
\({ T }_{ 2 }P={ T }_{ 2 }O+OP=D+x\)  and
\(\\ { T }_{ 1 }P={ T }_{ 1 }O-OP=D-x\)
\({ S }_{ 1 }P=\sqrt { \left( { S }_{ 1 }{ T }_{ 1 } \right) +\left( { PT }_{ 1 } \right) ^{ 2 } } =\sqrt { { D }^{ 2 }+\left( D-x \right) ^{ 2 } } \)
and  \({ S }_{ 2 }P=\sqrt { \left( { S }_{ 2 }{ T }_{ 2 } \right) +\left( { T }_{ 2 }P \right) ^{ 2 } } =\sqrt { { D }^{ 2 }+\left( D-x \right) ^{ 2 } } \)
The minima will occur when \({ S }_{ 2 }P-{ S }_{ 1 }P=\left( 2n-1 \right) \frac { \lambda }{ 2 } \)
i.e.\(\left[ { D }^{ 2 }+\left( D+x \right) ^{ 2 } \right] ^{ 1/2 }-\left[ { D }^{ 2 }+0 \right] ^{ 1/2 }=\frac { \lambda }{ 2 } \)
[for first minima, n = 1]  
if x = D, we can write, \(\left[ { D }^{ 2 }+4D^{ 2 } \right] ^{ 1/2 }-\left[ { D }^{ 2 }+0 \right] ^{ 1/2 }=\frac { \lambda }{ 2 } \)
\(\Rightarrow \left[ 5D^{ 2 } \right] ^{ 1/2 }-\left[ { D }^{ 2 } \right] ^{ 1/2 }=\frac { \lambda }{ 2 } \Rightarrow \sqrt { 5D-D } =\frac { \lambda }{ 2 } \)
\(\\ \Rightarrow D\left( \sqrt { 5-1 } \right) =\lambda /2orD=\frac { \lambda }{ 2\left( \sqrt { 5-1 } \right) } \)
putting \(\sqrt { 5 } =2.236\)
\(\\ \Rightarrow \sqrt { 5-1 } =2.236-1=1.236\)
\(\\ D=\frac { \lambda }{ 2\left( 1.236 \right) } =0.404\lambda \)

The least distance of discint vision of an average person. (i.e, D) is 25 cm, in order to view an object with magnificaton of 10. 
Here, v = D = 25 cm and u = f
But thr magnification, m = v/u = D/f
\(m=\frac { D }{ f } \Rightarrow f=\frac { D }{ m } =\frac { 25 }{ 10 } =2.5=0.025m\)
\( P=\frac { 1 }{ 0.025 } =40D \ [\therefore \ P=\frac { 1 }{ f } ]\)
This is the required power of lens.

Focal length for convex lens = f₁
Focal length for concave lens = -f₂
The equivalent focal length
\(\frac { 1 }{ F } =\frac { 1 }{ f_{ 1 } } +\frac { 1 }{ f_{ 2 } } \Rightarrow F=\frac { f_{ 1 }f_{ 2 } }{ f_{ 1 }-f_{ 2 } } \)

if there was no cut then object heigh 0.5 cm principal axis OO'
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }\)
\( \frac { 1 }{ v } =\frac { 1 }{ u } +\frac { 1 }{ f } \)
\(= \frac { 1 }{ -50 } +\frac { 1 }{ 25 } \)
\(= \frac { 1 }{ -50 } +\frac { 1 }{ 25 }\)
\( v=50 \ cm \ m=\frac { v }{ u } \ \Rightarrow \frac { -50 }{ 50 } =-1\)