Answer to Question #205112 in Optics for anuj

Question #205112

Suppose a 200 mm-focal length telephoto lens is being used to photograph mountains 10.0 km away.

(a) Where is the image? ______ m

(b) What is the height of the image of a 1000 m high cliff on one of the mountains? ______ m



1
Expert's answer
2021-06-11T11:20:42-0400

Lens Formula:\text{Lens Formula:}

1f=1u+1v;\frac{1}{f}=\frac{1}{u} +\frac{1}{v};

1v=1f1u;\frac{1}{v}=\frac{1}{f} -\frac{1}{u};

v=ufuf;v=\frac{u*f}{u-f};

f=200mm=0.2mf=200mm =0.2m

u=10km=10000mu=10km=10000m

v=100000.2100000.20.2mv=\frac{10000*0.2}{10000-0.2}\approx0.2m

Answer : (a) Where is the image? 0.2 m\text{Answer : (a) Where is the image? 0.2 m}


Magnification:\text{Magnification:}

m=Heigh of ImageHeigh of Object;m=\frac{\text{Heigh of Image}}{\text{Heigh of Object}}; ​

m=vum=\frac{v}{u}

Heigh of Image=Heigh of Objectvu\text{Heigh of Image}={\text{Heigh of Object}}*\frac{v}{u}

Heigh of Object=1000m{\text{Heigh of Object}}=1000m

Heigh of Image=10000.210000=0.02m\text{Heigh of Image}=1000*\frac{0.2}{10000}=0.02m

Answer: \text{Answer: }

What is the height of the image of a 1000 m high cliff on one of the mountains? 0.02m\text{What is the height of the image of a 1000 m high cliff on one of the mountains? 0.02m}



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