A 1.3 cm tall candle flame is 62cm from a lens with a focal lengthof 22 cm.

You are watching: What are the distance between the candle and the lens for each location?

**a) What is the image distance? in cmb) What is the height of the flame's image? Remember that anupright image has a positive height, whereas an inverted image hasa negative height. in cm**

The concept required to solve this problem is Lens equation.

Initially, use the lens formula to calculate the height of image formed. Finally use the magnification equation to solve for the focal length of the lens.

The lens equation is given as,

1s+1s′=1ffrac{1}{s} + frac{1}{{s'}} = frac{1}{f}s1+s′1=f1

Here, sss is the object distance, s′s's′ is the image distance, and fff is the focal length.

The magnification is the ratio of object height to image height. The equation of magnification is given as,

m=−s′sm = – frac{{s'}}{s}m=−ss′

Here, s′s's′ is the image position, and sss is the object position.

**(a)**

The expression to calculate the image distance is,

s′=fss−fs' = frac{{fs}}{{s – f}}s′=s−ffs

Substitute 22cm22{

m{ cm}}22cm for f and 62cm62{

m{ cm}}62cm for sss in the above equation s′=fss−fs' = frac{{fs}}{{s – f}}s′=s−ffs and calculate the image position.

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s′=(22cm)(62cm)(62cm)−(22cm)=34.1cmegin{array}{c}\s' = frac{{left( {22{

m{ cm}}}

ight)left( {62{

m{ cm}}}

ight)}}{{left( {62{

m{ cm}}}

ight) – left( {22{

m{ cm}}}

ight)}}\\ = 34.1{

m{ cm}}\end{array}s′=(62cm)−(22cm)(22cm)(62cm)=34.1cm

**(b)**

The equation of magnification is given as follows:

m=−s′sm = – frac{{s'}}{s}m=−ss′

Here, s′s's′ is the image position, and sss is the object position.

The magnification is also calculated as:

m=hihom = frac{{{h_i}}}{{{h_o}}}m=hohi

Here, hi{h_i}hi is the image height and ho{h_o}ho is the object height.

Equate both the equation of magnification.

hiho=−s′shi=ho(−s′s)egin{array}{c}\frac{{{h_i}}}{{{h_o}}} = – frac{{s'}}{s}\\{h_i} = {h_o}left( { – frac{{s'}}{s}}

ight)\end{array}hohi=−ss′hi=ho(−ss′)

Substitute 1.3 cm for ho{h_o}ho , 34.1 cm for s′s's′ and 62 cm for s in the above equation.

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hi=(1.3cm)(−(34.1cm)62cm)=−0.715cmegin{array}{c}\{h_i} = left( {{

m{1}}{

m{.3 cm}}}

ight)left( {frac{{ – left( {{

m{34}}{

m{.1 cm}}}

ight)}}{{{

m{62 cm}}}}}

ight)\\ = – 0.715{

m{ cm}}\end{array}hi=(1.3cm)(62cm−(34.1cm))=−0.715cm

Since, the image height is negative. Therefore, the image formed is inverted in nature.