Two thin lenses of power 4 D and 2 D are placed in contact then the combination is similar to

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A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the mircoscope have to be moved to focus on the needle again?

Case I: When tank is filled with water. Given, real depth = 12.5 cm;           apparent depth = 9.4 cm Now, using the formula, 

                    μ = real depthapparent depth

Refractive index,  μ = 12.59.4 = 1.33

Case II: When water in the tank is replaced by another liquid.

apparent depth = real depthμ 

i.e., apparent depth = 12.51.63 = 7.67 cm

Distance through which microscope has to be moved downward is = (9.4 – 7.67) cm = 1.73 cm.

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of a focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Here, the point P on the right of the lens acts as a virtual object.

Object distance, u = 12 cm

∴           1v =120+112      =3+560 =860

i.e.,           v = 60/8 = 7.5 cm. 

Image is at a distance of 7.5 cm to the right of the lens, where the beam converges. (b)Now,Focal length of concave lens, f = –16 cmObject distance, u = 12 cm 

⇒      v = 48 cm 

Hence, the image is at a distance of 48 cm to the right of the lens, where the beam would converge.