An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?

Engineering

Physics

Curved Surface Refraction

Question

5.6 m

3.2 m

7.2 m

2.4 m

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$\frac{1}{12} + \frac{1}{240} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{7}{80}$

$Δx due to slab = t (1 - \frac{1}{μ}) = \frac{1}{3} CM$

* New v should be $= 12 - \frac{1}{3} = \frac{35}{3} CM$

* $\frac{1}{u} = \frac{1}{v} \frac{1}{f} = \frac{3}{35} \frac{7}{80} = - \frac{1}{500}$

u = –5.6 m