A lens forms a real image of an object. The distance from the object to the lens is x cm and that from the lens to the image is y cm. the graph (Fig.) shows the variation of y with x. It can be deduced that the lens is

Question

A lens forms a real image of an object. The distance from the object to the lens is x cm and that from the lens to the image is y cm. the graph (Fig.) shows the variation of y with x. It can be deduced that the lens is (A) converging and of focal length 10 cm (B) converging and of focal length 20 cm. (C) converging and of focal length 40 cm (D) diverging and of focal length 20 cm

Tardigrade Admin · Accepted Answer

A diverging lens if ruled out because both

x

and

y

are positive values. Both

x

and

y

equal

20 c m

at their smallest sum, which occurs when

x + y = 40 c m = 4 f

∴ f = 10 c m

This indicates a converging lens of focal length.

= 10 c m

Q. A lens forms a real image of an object. The distance from the object to the lens is xcm and that from the lens to the image is ycm. the graph (Fig.) shows the variation of y with x. It can be deduced that the lens is

converging and of focal length 10 cm

converging and of focal length 20 cm.

converging and of focal length 40 cm

diverging and of focal length 20 cm

A diverging lens if ruled out because both x and y are positive values. Both x and y equal 20cm at their smallest sum, which occurs when x+y=40cm=4f ∴f=10cm This indicates a converging lens of focal length. =10cm

Q. A lens forms a real image of an object. The distance from the object to the lens is $x c m$ and that from the lens to the image is $y c m$ . the graph (Fig.) shows the variation of $y$ with $x$ . It can be deduced that the lens is

A diverging lens if ruled out because both $x$ and $y$ are positive values. Both $x$ and $y$ equal $20 c m$ at their smallest sum, which occurs when
$x + y = 40 c m = 4 f$
$∴ f = 10 c m$
This indicates a converging lens of focal length. $= 10 c m$