As the refractive lens index for red is less than that for blue light and a lens will be bent towards the axis for blue light as compared to red
\( (\frac { 1 }{ f } =(n_{ 21 }-1)\frac { 1 }{ R_{ 1 } } -\frac { 1 }{ R_{ 2 } } ,\ f_{ b }\)
focal length of blue light will be smaller than that of red.

the figure shows L in contact with a plane mirror M.P is the point object kept in fron of this combination at a distance 20 cm from it.
   
the image is conciding with the object itself the rays from the object and the image is formed at the focus of the lens.so focul length of the lens is 20 cm.

If the lower half of the concave mirror is blackened, then there is no change in the position of image, only intensity will get reduced.

Let \(\theta \) be the angle between the mirrors \({ M }_{ 1 }\) and \({ M }_{ 2 }\)
Since, rays LM and NS are parallel to each other

\(\angle LMN=\angle SNQ=2i\)
\( \Rightarrow \quad 2i_{ 1 }+2i=180^{ \circ }\)
\(\Rightarrow i_{ 1 }+i=90^{ \circ }...............(i)\)
\(Also \ in \ \triangle MBN,\)
\( \theta +\left( 90-i \right) +\left( 90-i \right) =180^{ \circ }\)
\(\Rightarrow \theta =i_{ 1 }+i..............(ii)\)
On comparing Eqs. (i) and (ii), we get
\(​​\theta =90^{ \circ }\)

The wavefront of light from a point source will be sherical and its intencity will decrease proportional to the square of distance.