The equiconvex lens has focal length f. If is cut perpendicular to the principal axis passin through optical centre, then focal length of each half is

Options

A. `f/2`
B. 2f
C. `(3f)/2`
D. f

Answer

B

Explanation

As, for equiconvex lens ` R _ 1 = R _ 2 = R ` ( say )
So, ` ( 1 ) / ( f ) = ( mu _ ("real " ) - 1 ) ( 2 ) / ( R ) " " `… ( i)
Now, if lens is cut along the line perpendicular to the principal axis as shown in the figure,

The new cut part of the lens has, ` R _ 1 = R and R _ 2 = oo `
Again by using the lens Maker's formula, focal length of the new part of the lens,
` ( 1 ) /( f' ) = ( mu _ ("real" ) - 1) [ ( 1 ) /( R) - ( - ( 1 ) /( oo) ) ] `
` rArr ( 1 ) / ( f) = ( mu _ ( "real " ) - 1 ) [ ( 1 ) /( R) ]" "...(ii) `
So, from the Eqs. (i ) and (ii) , we get
` f ' = 2 f `