1.

In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The magnification of the telescope is :

Answer»

`L/I`
`L/I+1`
`L/I-1`
`(L+1)/(L-1)`

Solution :
MAGNIFICATION of TELESCOPE, m = `(f_o)/(f_e)`
But, magnification m = `-("HEIGHT of image")/("height of OBJECT")`
`v/u=-1/L`
`therefore (f_e)/(f_e-(f_o+f_e))=-I/L`
`therefore (f_e)/(-f_o)=-I/L`
`(f_o)/(f_e)=L/T`
`therefore` magnification `m=L/I`


Discussion

No Comment Found

Related InterviewSolutions