Question

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

the:
image distance from the lens? -
the magnification
nature of the image
.
.

Answer 1

» Image distance :

`{ \tt{ (1)/(v) + (1)/(u) = (1)/(f) }} \\`

• v is image distance
• u is object distance, u is 10 cm
• f is focal length, f is 5 cm

`{ \tt{ (1)/(v) + (1)/(10) = (1)/(5) }} \\ \\ { \tt{ (1)/(v) = (1)/(10) }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}`

» Magnification :

• Let's derive this formula from the lens formula:

`{ \tt{ (1)/(v) + (1)/(u) = (1)/(f) }} \\`

» Multiply throughout by fv

`{ \tt{fv( (1)/(v) + (1)/(u) ) = fv( (1)/(f) )}} \\ \\ { \tt{ (fv)/(v) + (fv)/(u) = (fv)/(f) }} \\ \\ { \tt{f + f( (v)/(u) ) = v}}`

• But we know that, v/u is M

`{ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = (v)/(f) }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = (v)/(f) - 1 }}}}`

• v is image distance, v is 10 cm
• f is focal length, f is 5 cm
• M is magnification.

`{ \tt{M = (10)/(5) - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}`

» Nature of Image :

• Image is magnified
• Image is erect or upright
• Image is inverted
• Image distance is identical to object distance.