An object of height 6 cm is placed perpendicular to the principal axis of a concave axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm.

Explanation

Height of the object, `h_(1) = 6 cm`
Focal length of the concave mirror, `f= -5 cm`
Position of the image, `v= ?" Size of the image , " h_(2)=?`
Object distance , `u= -10 cm`
According to lens formula:
`(1)/(v)-(1)/(u)=(1)/(f) rArr (1)/(v)-(1)/(-10)=(1)/(-5)rArr (1)/(v)+(1)/(10)=(-1)/(5)`
`rArr (1)/(v)= (-1)/(5)-(1)/(10)=(-2-1)/(10)=(-3)/(10) therefore v = (-10)/(3) = -3.3 cm`
`(h_(2))/(h_(1)) = (v)/(u) rArr (h_(2))/(6)= (cancel(-)(10)/(3))/(cancel(-)10)`
`rArr (h_(2))/(6)=(10)/(3xx10)=(1)/(3) rArr h_(2)=(6)/(3) therefore h_(2) = + 2 cm`
Thus the image is formed at a distance of 3.3 cm from the concave lens. The negative (-) sign for image distance shows that the image is formed on the left side of the concave lens (i.e., virtual). The size of the image is 2 cm and the positive (+) sign for hand image shows that the image is erect.
Thus a virtual, erect, diminished image is formed on the same side of the object (i.e., left side).