Question

An object 2cm high is placed at a distance of 64cm from a white screen. On placing a convex lens at a distance of 32cm from the object it is found that a distinct image of the object is formed on the screen.what is the focal length of the convex lens and the size of image formed on the screen

Answer 1

Determination of focal length of the lens:

From the lens formula, 
1f=1v-1u     =164-32-1-32     =132+132     =232f=322  =16 cm

Determination of the size of the image:

From the formula of magnification,

m=h2h1=vuh2=vu×h1    =32-32×2    = -2 cm

Here, the negative sign of image size represent that the image is inverted.

Ray diagram to show the formation of lens:

diagram is on the description...

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Answer 2

Answer:

$f=16$ , $h=-2$

Explanation:

Concept

• The focal length is the distance between a convex lens or a concave mirror and the focal point of a lens or mirror. It is the point at which two parallel light rays meet or converge. Depending on the nature of the lens and mirror, the focal length varies with the sign (positive or negative) (concave or convex).

Given

Height(H)=2 Cm

Distance=64 Cm from a white screen

Distance=32 Cm convex lens facing

Find

Focal Length of a convex lens

Size of the image formed on the screen

Solution

$\frac{1}{v}-\frac{1}{u} =\frac{1}{f}$

$\frac{1}{32}-\frac{1}{-32} =\frac{1}{f} \quad \text { or } \quad \frac{1}{32}+\frac{1}{32}=\frac{1}{f}$

$f=16$

$\Rightarrow \quad \text { or } \frac{h^{\prime}}{h} =\frac{v}{u}$

$\Rightarrow \quad h^{\prime} =\frac{v}{u} \times h$

$\therefore \quad h^{\prime} =\frac{32}{-32} \times 2=-2 \mathrm{~cm}$

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