Science — CBSE Class 10 board question

(i)

When a ray of light passes through a rectangular glass slab, it refracts at two parallel faces — AB (air to glass) and CD (glass to air). The bending at these two faces is equal and opposite because the faces are parallel to each other. Hence, the emergent ray comes out parallel to the incident ray, though it is laterally displaced.

Ray Diagram:

```
E N
\ |
\i |
----O-------+---- AB (air-glass)
|r |
OO'(refracted ray)
----O'------+---- CD (glass-air)
| |
\ M'
H (emergent, parallel to EO)
```

When light falls normally on a face: The angle of incidence = 0°, so the ray passes straight through without any bending (no refraction). It continues in the same direction.

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(ii)

Given: $u = -30$ cm, $f = -20$ cm (concave lens)

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{-30}$$

$$\frac{1}{v} = \frac{-3 - 2}{60} = \frac{-5}{60} = \frac{-1}{12}$$

$$v = -12 \text{ cm}$$

The image is formed 12 cm in front of the lens (on the same side as the object). It is virtual and erect.

Source: Chapter 9, Section 9.3.1 (Refraction through a Rectangular Glass Slab) and Sign Convention for Spherical Lenses

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• (i) The key reason is "equal and opposite bending at two parallel faces" — state this clearly. Examiners award marks for the diagram showing incident ray, refracted ray inside slab, and emergent ray parallel to incident. For normal incidence, state that angle of incidence = 0°, so no bending occurs — the ray goes straight through.
• (ii) Always write the lens formula first, substitute with correct signs (both u and f are negative for a concave lens with real object), and state the nature of the image. A negative v confirms the image is on the same side as the object — virtual and erect, consistent with Table 9.5.