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Q.A point source in air is kept 24 cm in front of a concave spherical glass surface (aμg=1.5)({}_a\mu_g = 1.5) and radius of curvature 60 cm. Find the nature of the image formed and its distance from the point source.

CBSECBSE Class XII Board 2022Subjective· 3mImportance★★★★★
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Using refraction at a single spherical surface with the concave surface taken as R=−60 cmR=-60\ \text{cm}, the image is virtual, erect and diminished, formed 30 cm30\ \text{cm} from the surface on the same side as the source — i.e. 6 cm6\ \text{cm} from the point source.

This is refraction at one spherical surface (air →\to glass), governed by

μ2v−μ1u=μ2−μ1R.\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}.

Sign convention (light from source into glass):

  • μ1=1\mu_1=1 (air), μ2=1.5\mu_2=1.5 (glass)
  • u=−24 cmu=-24\ \text{cm} (real object)
  • The surface is concave towards the source, so its centre of curvature lies on the incident side: R=−60 cmR=-60\ \text{cm}.

Substituting:

1.5v−1−24=1.5−1−60\frac{1.5}{v}-\frac{1}{-24}=\frac{1.5-1}{-60}

1.5v+124=−1120\frac{1.5}{v}+\frac{1}{24}=-\frac{1}{120}

1.5v=−1120−5120=−6120=−120\frac{1.5}{v}=-\frac{1}{120}-\frac{5}{120}=-\frac{6}{120}=-\frac{1}{20}

v=1.5×(−20)=−30 cm.v=1.5\times(-20)=-30\ \text{cm}.

The negative vv means the image lies on the same side as the object (in air): it is virtual and erect. The magnification is …

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