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Engineering
5 months, 1 week ago

A point object is placed in front of a lens at distance 10 cm from convex lens Lens produces a sharp image atscreen located at 10 cm from

A point object is placed in front of a lens at distance 10 cm from convex lens. Lens produces a sharp image at screen located at 10 cm from lens. Now a glass slab of refractive index 1.5 & thickness 1.5 cm is placed in contact with lens at its left side. Now in order to get sharp image again on the screen by what distance screen should be shifted.

(1) 0.5 (2) 0.55 (3) 0.4 (4) None

Answers (1)
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A Avinash Dongre
Answered 3 months, 1 week ago

 

F = \frac{10}{2} = 5 cm

Due to glass plate shift

s = t \left ( 1 - \frac{1}{ \mu } \right )

= 1.5 \left ( 1 - \frac{2}{ 3 } \right )

= 0.5

New u =9.5 cm

Using lens formula

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

= \frac{1}{5} + \frac{1}{-9.5}

\frac{1}{v} = \frac{+4.5}{+9.5\times 5}

v = 10.55 cm

Shift  is 0.55 cm away from the lens.

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