Answer to Question #114551 in Optics for Jason

Question

Consider the thin lenses in the drawing (n= 1.55). Suppose that for each of them R1= 25 cm and R2= 60 cm. 1) Find the focal lengths for each lens, and use their signs to obtain the location of the image of a distant source. 2) For each lens, is the image upright or inverted, virtual or real? 3) Calculate the magnification, and check that it agrees with your answer to the previous question.

Answer 1

As per the question,

The diagram mentioned in the question is not attached, let the position of object is u and the position of the image is v,

let the focal length of the object is f.

Refractive index of the lens $(\mu)=1.55$

Radius of curvature of the lens $(R_1)=25cm$

and radius of the second curvature $(R_2)=60cm$

a) We know that,

$\dfrac{1}{f}=(\mu -1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

Now substituting the values,

$\Rightarrow \dfrac{1}{f}=(1.55-1)(\dfrac{1}{25}-\dfrac{1}{60})=0.55\times (\dfrac{7}{300})$

$\Rightarrow f= \dfrac{300}{3.85}=77.92cm$

b) Now we will apply the lens formula

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

we will substitute the given value as per the image, which is not attach with this question, then we will get it.

if magnification m is positive then it is real if it is negative then it will be virtual. if m = hi/ho is positive then it will be upright if it is negative then image will be inverted.