D

What is the result of this setup?

2 viewed last edited 1 year ago
Anonymous
0

Two plano-concave lenses of glass of refractive index 1.5 have radii of curvature 25 cm and 20 cm. They are placed in contact with their curved surface towards each other and the space between them is filled with liquid of refractive index 4/3. What does this setup act as?


**By that I mean (concave/convex) and the focal length of it...

Krishna
0

Step 1: Understand the question and underline the main points

            NOTE:  Refractive index   μ = 1.5

                        First lens radius of curvature R_1= 25 cm

                        Second lens radius of curvature R_2 = 20 cm

            The space between the lenses is filled with liquid of refractive index (μ) = \frac{4}{3}.


Step 2: Construct an imaginary figure by using the given data

We know that

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}......................(1)

        f = focal length 


Step 3: Recall the lens makers formula

            Know about lens makers formula

            LINK: https://byjus.com/physics/derivation-of-lens-maker-formula/


      FORMULA: \frac{1}{f}   = ( μ - 1)  [math] [\frac{1}{R_1} - \frac{1}{R_2}] [/math]


\frac{1}{f_1}  = ( 1.5 - 1)  [math] [\frac{1}{\infty} - \frac{1}{25}] = -12.5 [/math]

\frac{1}{f_2}= (  \frac{4}{3}. - 1)  [math] [\frac{1}{25} - \frac{1}{-20}] =0.03 [/math]

\frac{1}{f_3}  = ( 1.5 - 1)  [math] [\frac{1}{-20} - \frac{1}{\infty}] = - 0.025 [/math]  


From equation (1)

\frac{1}{F} = -12.5 + 0.03 + -0.025 = -12.495

                     F = 0.080