NEET Practice Questions (MCQs) with Answers & Solutions

$ use , \therefore {1 \over f } = (n-1) \left( {1 \over R_1} + {1 \over R_2 } \right) take R_1 = R_1 , R_2 = -2R $ R = 4.5 cm

$ here , n = { sin \over sin r } = { sin (i) \over sin ({i \over 2} ) } = { 2 sin ({i \over 2 } ) . cos ({i \over 2} ) \over sin ({i \over 2 } ) } $ $ \therefore n = cos \left( { i \over 2 } \right) $ $ \therefore {n \over 2 } = cos \left( { i \over 2 } \right) $ $ \therefore i = 2 cos^{-1} \left( { n \over 2 } \right) $

For two plano-convex lenses with the same radius of curvature R and refractive index n = 1.5, when placed in contact with each other, the focal length (f) is given by:

$$ \frac{1}{f} = (n-1) \left( \frac{1}{R} - \frac{1}{-R} \right) $$

Simplifying this, we get:

$$ \frac{1}{f} = (1.5-1) \left( \frac{2}{R} \right) $$

$$ \frac{1}{f} = \frac{1}{R} $$

Therefore, the focal length f = R. Hence, the correct answer is 'in contact with each other'.

wavelength maximum, Scattering is minimum

$ here, n = { sin \left( { A + A \over 2 } \right) \over sin \left( A / 2 \right) } ={ 2sin \left( { A \over 2 } \right). cos \left( { A \over 2 } \right) \over sin \left( { A \over 2 } \right) }$ $ { 3 \over 4 } = 2 cos \left( { A \over 2 } \right) ,{ A \over 2 } = cos ^{-1} (0.75) = 41 ^\circ , \therefore A = 82 ^\circ $

$ here , \delta m = r \;and \;\delta - i _1 + i_2 - ( r_1 -r_2 ) $ $ when \delta = \delta m \;then i_1 = i_2 = i , r_1 = r_2 = r $ $ \therefore \delta m = 2i -2r = 2nr - 2r $ $ \left( \therefore n = { sin i \over sin p } = { i \over r } \therefore i = nr \right) $ $ = 2r (n-1) = 2r\left( {3 \over2} -1 \right) $ $ \therefore \delta m = r $

Here magnifying power is 10 there it can be seen 10 times near.

The magnifying power of a microscope is inversely proportional to the focal length of the objective lens. When the length of the microscope tube increases, the effective focal length increases, leading to a decrease in magnifying power. Therefore, the correct answer is 'decreases'.